The definition of a circle and the derivation of its standard equation from the distance formula, graphing a circle from a rearranged equation, writing an equation from a point on the circle, finding tangent lines using the perpendicular radius, and using circular inequalities to model coverage regions.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 9 — Quadratic Relations and Conic Sections
Graph and Write Equations of Circles
Objectives
Five outcomes. One distance, held fixed, and the distance formula does the rest.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.3 Graph and Write Equations of Circles §9.3, pp. 626-633 — the lesson these objectives are drawn from
Warm-up
Lesson 9.1 gave you the distance formula and Lesson 9.2 defined a curve by distances.
Discussion prompt
Which points sit exactly 5 units from the origin? Write down four of them, then say what shape they all lie on.
Hint: Try 3 and 4 as coordinates.
Answer:
\[ (5,0), \; (0,5), \; (3,4), \; (-4,3): \quad \text{all at distance } 5 \]
\[ \sqrt{x^2+y^2} = 5 \;\Longrightarrow\; x^2+y^2 = 25 \]
They lie on a circle of radius 5. Squaring both sides cleared the radical, and that one move turns the distance formula into the equation of every circle in this lesson.
Concept
A circle is the set of points a fixed distance from a fixed centre. Writing that distance with the distance formula and squaring both sides gives the standard equation for a circle centred at the origin.
circle — The set of all points in a plane equidistant from a fixed point called the centre. The common distance is the radius.
\[ \sqrt{x^2+y^2} = r \;\Longrightarrow\; x^2+y^2 = r^2 \]
The equation has r squared rather than r, which is why an ugly radical radius still gives a tidy equation. It also means the radius is recovered by taking a square root.
Figure (svg): Two columns comparing the parabola's definition with the circle's
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.3 Graph and Write Equations of Circles §9.3, pp. 626-626
Section
Section 1
Concept
For a circle centred at the origin, every point on it satisfies the distance formula with the distance equal to r. Squaring both sides removes the radical and leaves the sum of two squares equal to r squared.
\[ x^2+y^2 = r^2 \]
Both variables are squared and both carry the same coefficient. That is what distinguishes a circle's equation from an ellipse's in Lesson 9.4, where the coefficients differ.
Figure (svg): A circle centred at the origin, with the distance formula turned into its equation
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.3 Graph and Write Equations of Circles §9.3, pp. 626-626 — Standard Equation of a Circle with Center at the Origin
Picture it
The derivation in two lines.
Figure (svg): A circle centred at the origin, with the distance formula turned into its equation
The dashed legs are the same right triangle as in Lesson 9.1. Only the fixed hypotenuse is new, and squaring makes the equation polynomial.
Worked example
The key concept's derivation.
\[ \text{Derive } x^2+y^2=r^2 \text{ from the definition of a circle of radius } r \text{ at the origin.} \]
State the definition
Why: Every point is r units from the centre.
\[ \text{distance } = r \]
Write the distance formula
Why: From the origin to a general point.
\[ \sqrt{(x - 0) ^{2} + (y - 0) ^{2}} = r \]
Square both sides
Why: This is legitimate since both sides are non-negative.
\[ (x - 0) ^{2} + (y - 0) ^{2} = r ^{2} \]
Simplify
Why: Subtracting zero changes nothing.
\[ x ^{2} + y ^{2} = r ^{2} \]
Figure (svg): A circle centred at the origin, with the distance formula turned into its equation
\[ x^2+y^2 = r^2 \]
Verify: test a point
Why: For r equal to 5, the point 3 comma 4 gives 9 plus 16, which is 25, and 5 squared is 25. So that point is on the circle. Squaring both sides was safe here because a distance and a radius are both non-negative, so no extraneous solutions can appear — unlike the radical equations of Lesson 6.6.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.3 Graph and Write Equations of Circles §9.3, pp. 626-626
Fill the middle
Example 1.
Fill in the blanks
x^2+y^2 = 36 \;\Longrightarrow\; r = \sqrt6 = ___
Why: The constant is r squared, so the radius is its square root, 6. Only the positive root is used, since a radius is a length.
Worked example
What makes an equation a circle rather than something else.
\[ \text{Which are circles centred at the origin? } x^2+y^2=36, \; x^2-y^2=36, \; 2x^2+2y^2=18, \; x^2+y^2=-9. \]
First: both squares, same coefficient, positive constant
Why: Radius 6.
Second: the signs differ
Why: A difference of squares is a hyperbola, not a circle.
Third: divide through by 2
Why: X squared plus y squared equals 9.
\[ a\text{ circle of radius } 3 \]
Fourth: the constant is negative
Why: No sum of two squares can be negative.
Figure (svg): The solution to Worked example recognise a circle's equation shown as a ladder of expressions, one row per algebraic move
\[ r = 6; \; \text{not a circle}; \; r = 3; \; \text{no points} \]
Verify: check the third by testing a point
Why: Dividing 2x squared plus 2y squared equals 18 by 2 gives radius 3, so 3 comma 0 should work: 2 times 9 plus 0 is 18. It does. Equal coefficients on the two squares are what allow the division to leave a clean sum, and unequal ones would give the ellipse of Lesson 9.4.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.3 Graph and Write Equations of Circles §9.3, pp. 626-626
Trap
\[ x^2+y^2 = 36 \]
Take the radius as 36
Why: The number on the right is read as the radius directly.
\[ r = 36 \quad \text{(wrong)} \]
Then the point 36 comma 0 would be on the circle, but 36 squared is 1296, not 36.
\[ r^2 = 36 \;\Longrightarrow\; r = 6 \]
The constant is r SQUARED, so take a square root
Why: The equation was built by squaring the distance, so it must be unsquared to recover a length.
\[ 6^2 + 0^2 = 36 \quad \checkmark \]
Only the positive root is taken, since a radius is a length. This is the same convention as the positive base in Lesson 7.7's exponential fitting.
Sorting
Both squares, same sign, same coefficient.
Sort into buckets
Sort each equation.
The last item was disguised: rearranging it gives x squared plus y squared equals 49. Getting to standard form before judging is always the first move.
Matching
The constant is r squared.
Match the pairs
Why: Two of the four radii are irrational, which is perfectly ordinary — and notice how much tidier the equations are than the radii. That is the advantage of writing r squared rather than r.
Prediction
Commit before reasoning.
Predict first
Lesson 6.6 warned that squaring both sides can create extraneous solutions. Why is it safe in this derivation?
Correct: Because both sides are non-negative, so squaring is reversible here.
\[ a, b \geq 0 \;\Longrightarrow\; (a = b \;\Longleftrightarrow\; a^2 = b^2) \]
Why: A distance is never negative and a radius is never negative, so no information is lost: the squared equation holds exactly when the original does. Squaring goes wrong only when it could equate a positive with a negative, which cannot happen between two lengths. That is why the circle's equation is genuinely equivalent to its definition, rather than merely implied by it.
Section
Section 2
Concept
Move both squared terms to one side, read r squared from the constant, take its square root, and plot the four points on the axes at that distance from the origin. Then sketch the curve through them.
\[ y^2 = -x^2+36 \;\Longrightarrow\; x^2+y^2=36, \; r=6 \]
The axis points are the easiest to plot because one coordinate is zero there. Adding a point like the one where both coordinates are r over the square root of 2 improves the sketch if needed.
Figure (svg): A circle graphed from a rearranged equation, with four convenient points marked
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.3 Graph and Write Equations of Circles §9.3, pp. 626-626 — Graph an equation of a circle
Picture it
Example 1, after rearranging into standard form.
Figure (svg): A circle graphed from a rearranged equation, with four convenient points marked
The circle passes through 6 and negative 6 on both axes. Nothing else needs computing, since the radius is the same in every direction.
Worked example
Example 1.
\[ \text{Graph } y^2 = -x^2+36 \text{ and identify the radius.} \]
Rewrite in standard form
Why: Add x squared to both sides.
\[ x ^{2} + y ^{2} = 36 \]
Identify the centre
Why: There are no shifts, so it is the origin.
\[ (0, 0) \]
Find the radius
Why: The square root of 36.
\[ r = 6 \]
Plot and sketch
Why: The four axis points at distance 6.
\[ (6, 0), (-6, 0), (0, 6), (0, -6) \]
Figure (svg): A circle graphed from a rearranged equation, with four convenient points marked
\[ x^2+y^2 = 36, \quad r = 6 \]
Verify: test a non-axis point
Why: The point where x is 3 and y is the square root of 27 should be on the circle: 9 plus 27 is 36. Its distance from the origin is the square root of 36, which is 6 — the radius, as required. Every point of the curve passes the same test, which is what the equation asserts.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.3 Graph and Write Equations of Circles §9.3, pp. 626-626
Fill the middle
Example 1.
Fill in the blanks
y^2 = -x^2+36 \;\Longrightarrow\; x^2+y^2 = 36
Why: Adding x squared to both sides puts both squares on the left and leaves 36 on the right, so r squared is 36 and the radius is 6.
Worked example
Guided Practice 1 to 3.
\[ \text{Graph } x^2+y^2=9, \; y^2=-x^2+49, \; x^2-18=-y^2 \text{ and give each radius.} \]
First: already standard
Why: The square root of 9.
\[ r = 3 \]
Second: add x squared
Why: X squared plus y squared equals 49.
\[ r = 7 \]
Third: add y squared and 18
Why: X squared plus y squared equals 18.
\[ r = \sqrt{18} \]
Third: simplify the radical
Why: Eighteen is 9 times 2.
\[ r = 3 \sqrt{2},\text{ about } 4.24 \]
Figure (svg): The solution to Worked example three more circles shown as a ladder of expressions, one row per algebraic move
\[ r = 3, \; 7, \; 3\sqrt{2} \approx 4.24 \]
Verify: check the third's rearrangement
Why: Starting from x squared minus 18 equals negative y squared, adding y squared gives x squared plus y squared minus 18 equals 0, and adding 18 gives 18 on the right. The radius is then root 18, or 3 root 2, which sits between 4 and 5 — so the circle passes just outside the point 4 comma 0 and just inside 5 comma 0.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.3 Graph and Write Equations of Circles §9.3, pp. 627-627
Error analysis
A student reads the radius from an unrearranged equation.
Annotate
On: \( y^2 = -x^2+36 \;\Longrightarrow\; r = \sqrt{-36} \text{, undefined} \)
Standard form first, every time. Only once both squared terms sit together on one side does the constant mean r squared.
Ranking
Graphing a circle from any arrangement.
Put in order
Why: Step one cannot be skipped, since the constant only means r squared once the equation is in standard form. Everything after that is mechanical.
Sorting
Compare the sum of squares with 36.
Sort into buckets
Sort each point for the circle x squared plus y squared equals 36.
This three-way test is exactly the inequality idea of the last section, and it costs one addition per point.
Prediction
Commit before reasoning.
Predict first
Why are four points enough to sketch a circle when a parabola needed a table?
Correct: Because the radius is the same in every direction, so the shape is completely determined.
\[ \text{centre} + r \;\Longrightarrow\; \text{the whole circle} \]
Why: A circle has no varying width or curvature to discover — knowing the centre and one distance fixes every point of it. In fact a single point would do, since it gives the radius. The four axis points are plotted only to guide the hand, whereas a parabola's shape genuinely changes along the curve and needs sampling to reveal it.
Section
Section 3
Concept
If a circle is centred at the origin and a point on it is known, the radius is the distance from the origin to that point. Squaring it gives the constant, so the radical never survives into the equation.
\[ r = \sqrt{29} \;\Longrightarrow\; x^2+y^2 = 29 \]
In fact the squaring can be skipped entirely: r squared is the sum of the squared coordinates, so the constant is available without ever taking a root.
Figure (svg): A circle whose radius is found as the distance from the origin to a given point
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.3 Graph and Write Equations of Circles §9.3, pp. 627-627 — Write an equation of a circle
Picture it
Example 2: the point where x is 2 and y is negative 5.
Figure (svg): A circle whose radius is found as the distance from the origin to a given point
The radius is the square root of 29, but the equation is simply x squared plus y squared equals 29 — the radical vanishes on squaring.
Worked example
Example 2.
\[ \text{The point } (2,-5) \text{ lies on a circle centred at the origin. Write its equation.} \]
Find the radius
Why: The distance from the origin to that point.
\[ \sqrt{4 + 25} = \sqrt{29} \]
Write the standard form
Why: With r equal to the square root of 29.
\[ x ^{2} + y ^{2} = (\sqrt{29}) ^{2} \]
Simplify
Why: Squaring undoes the square root.
\[ x ^{2} + y ^{2} = 29 \]
Notice the shortcut
Why: The constant is just the sum of the squared coordinates.
\[ 4 + 25 = 29 \]
Figure (svg): A circle whose radius is found as the distance from the origin to a given point
\[ x^2+y^2 = 29 \]
Verify: substitute the given point
Why: Two squared plus negative 5 squared is 4 plus 25, which is 29 — so the point is on the circle, as required. The shortcut in the last step is worth adopting: computing r squared directly avoids taking a square root only to square it again.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.3 Graph and Write Equations of Circles §9.3, pp. 627-627
Fill the middle
Example 2.
Fill in the blanks
r^2 = 2^2+(-5)^2 = 29
Why: Four plus 25 is 29, which is the constant of the equation. Taking a square root and then squaring it again would give the same number by a longer route.
Worked example
Guided Practice 4.
\[ \text{Write the equation of the circle through } (5,-1) \text{ with centre the origin.} \]
Square the coordinates
Why: Twenty-five and 1.
\[ 25\text{ and } 1 \]
Add them
Why: That sum is r squared.
\[ 26 \]
Write the equation
Why: No square root is ever needed.
\[ x ^{2} + y ^{2} = 26 \]
State the radius if asked
Why: The square root of 26 has no square factor.
\[ r = \sqrt{26} \]
Figure (svg): The solution to Worked example one more equation shown as a ladder of expressions, one row per algebraic move
\[ x^2+y^2 = 26 \]
Verify: check the point and estimate the radius
Why: Twenty-five plus 1 is 26, so the point is on the circle. And root 26 is a little over 5.09, which makes sense: the point is 5 across and 1 down, so its distance must be slightly more than 5. Estimating the radius is a quick way to catch a mis-added constant.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.3 Graph and Write Equations of Circles §9.3, pp. 627-627
Trap
\[ r = \sqrt{29} \]
Substitute the radius into the equation
Why: The letter r is replaced by its value directly.
\[ x^2+y^2 = \sqrt{29} \quad \text{(wrong)} \]
Testing the given point gives 29 on the left and about 5.39 on the right. The standard form has r SQUARED on the right.
\[ x^2+y^2 = r^2 = (\sqrt{29})^2 = 29 \]
Square the radius before substituting
Why: The right-hand side is r squared, not r.
\[ 2^2+(-5)^2 = 29 \quad \checkmark \]
Better still, compute the sum of the squared coordinates directly and skip the root entirely — the answer was 29 from the very first line.
Matching
The constant is the sum of the squared coordinates.
Match the pairs
Why: Only the third gives a whole-number radius, since 25 is a perfect square. The others have irrational radii but perfectly tidy equations, which is the point of writing r squared.
Two truths and a lie
All three are about writing a circle's equation.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. The radius root 29 is irrational, yet the equation is simply x squared plus y squared equals 29. Because the standard form uses r squared, radicals disappear rather than propagate — which is why the constant is easier to work with than the radius itself.
Prediction
Commit before reasoning.
Predict first
The points (2, -5) and (-5, 2) are both used to write a circle centred at the origin. What happens?
Correct: The same circle, since both have the same distance from the origin.
\[ 2^2+(-5)^2 = (-5)^2+2^2 = 29 \]
Why: Both give 4 plus 25, or 29, so both produce x squared plus y squared equals 29. Only the DISTANCE from the centre matters, not the direction, which is precisely what a circle's definition says. Any of the infinitely many points on that circle would give the same equation, and that is the sense in which one point determines the whole curve here.
Section
Section 4
Concept
A line tangent to a circle touches it at one point and is perpendicular to the radius drawn to that point. So its slope is the negative reciprocal of the radius's slope, and point-slope form finishes the job.
\[ m_{\text{radius}} = -3 \;\Longrightarrow\; m_{\text{tangent}} = \tfrac{1}{3} \]
Every ingredient is from Lesson 9.1: a slope between two points, a negative reciprocal, and point-slope form. Only the geometric fact about tangency is new.
Figure (svg): A circle with a tangent line drawn perpendicular to the radius at the point of contact
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.3 Graph and Write Equations of Circles §9.3, pp. 627-627 — Tangent to a circle
Picture it
Example 3: the tangent at a given point on a circle.
Figure (svg): A circle with a tangent line drawn perpendicular to the radius at the point of contact
The radius has slope negative 3 and the tangent one third, and their product is negative 1. The tangent touches the circle only at that one point.
Worked example
Example 3, a multiple-choice item.
\[ \text{Find the tangent to } x^2+y^2=10 \text{ at } (-1,3). \]
Find the radius's slope
Why: From the origin to the point of tangency.
\[ \frac{3}{-1} = -3 \]
Take the negative reciprocal
Why: The tangent is perpendicular to the radius.
\[ \frac{1}{3} \]
Write point-slope form
Why: Through negative 1 comma 3 with slope one third.
\[ y - 3 = (\frac{1}{3}) (x + 1) \]
Simplify
Why: Distribute and add 3.
\[ y = \frac{x}{3} + \frac{10}{3} \]
Figure (svg): A circle with a tangent line drawn perpendicular to the radius at the point of contact
\[ y = \tfrac{1}{3}x+\tfrac{10}{3} \]
Verify: check the point and the sketch
Why: At x equal to negative 1 the line gives negative one third plus ten thirds, which is 3 — the point of tangency. And a sketch shows the tangent there must rise from left to right, so a positive slope is right; the book uses exactly that observation to eliminate the choice with slope negative one third.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.3 Graph and Write Equations of Circles §9.3, pp. 627-627
Fill the middle
Example 3.
Fill in the blanks
m_3} = -3 \;\Longrightarrow\; m____} = \frac______}
Why: The negative reciprocal of negative 3 is positive one third. Their product is negative 1, which is the perpendicularity condition from Lesson 2.2.
Worked example
Guided Practice 5.
\[ \text{Find the tangent to } x^2+y^2=37 \text{ at } (6,1). \]
Check the point is on the circle
Why: Thirty-six plus 1 is 37.
Find the radius's slope
Why: One over 6.
\[ \frac{1}{6} \]
Take the negative reciprocal
Why: The negative of the reciprocal of one sixth.
\[ -6 \]
Write and simplify
Why: Through 6 comma 1 with slope negative 6.
\[ y = -6 x + 37 \]
Figure (svg): The solution to Worked example a second tangent shown as a ladder of expressions, one row per algebraic move
\[ y = -6x+37 \]
Verify: check the point and the product of slopes
Why: At x equal to 6 the line gives negative 36 plus 37, which is 1 — correct. And one sixth times negative 6 is negative 1, confirming perpendicularity. Both checks together confirm the line touches at the right place and in the right direction.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.3 Graph and Write Equations of Circles §9.3, pp. 627-627
Error analysis
A student writes the tangent line to a circle at a point where the radius has slope negative 3.
Annotate
On: \( y-3 = -3(x+1) \;\Longrightarrow\; y = -3x \)
The line found is the radius extended, which meets the circle at two points rather than touching at one. The negative reciprocal is what turns it into a tangent.
Ranking
Finding a tangent line to a circle.
Put in order
Why: Step one is worth ten seconds: if the given point is not on the circle, there is no tangent line at it and the whole calculation is meaningless. The rest is Lesson 9.1's machinery applied to a new geometric fact.
Sorting
Count how many points a line shares with the circle.
Sort into buckets
Sort each line's relationship with the circle x squared plus y squared equals 10.
Any line through the centre is automatically a secant, which is why using the radius's own slope can never produce a tangent.
Prediction
Commit before reasoning.
Predict first
Why must a tangent line be perpendicular to the radius at the point of contact?
Correct: Because the point of contact is the nearest point of the line to the centre, and the nearest point is reached perpendicularly.
\[ \text{shortest distance to a line} = \text{the perpendicular one} \]
Why: A tangent touches without entering, so every other point of it lies farther from the centre than r, making the contact point the closest one on the line. And the shortest segment from a point to a line is always the perpendicular one — the same fact used in Lesson 9.2 for measuring to a directrix. So the radius to the contact point must meet the line at a right angle.
Section
Section 5
Concept
The equation describes the boundary only. Replacing the equals sign with a less-than describes everything inside the circle, and a greater-than everything outside. That makes a circle the natural model for a coverage region.
\[ x^2+y^2 < r^2 \text{ inside}; \quad x^2+y^2 > r^2 \text{ outside} \]
Testing a point costs two squarings and an addition, with no square roots at all — which is why coverage checks are done with the squared inequality rather than by computing distances.
Figure (svg): A coverage circle around a phone tower, with a point inside it and the exit point to the south
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.3 Graph and Write Equations of Circles §9.3, pp. 628-628 — Circles and inequalities
Picture it
Examples 4 and 5: a 10 mile radius, a point inside it, and the exit point south.
Figure (svg): A coverage circle around a phone tower, with a point inside it and the exit point to the south
The flat tyre at 4 comma 9 gives 97, inside 100. Driving south the coverage ends at about 4 comma negative 9.2, giving 18.2 miles of range.
Worked example
Example 4.
\[ \text{A tower covers a } 10 \text{ mile radius. Is the point } 4 \text{ east and } 9 \text{ north in range?} \]
Write the inequality
Why: Everything within 10 miles of the tower at the origin.
\[ x ^{2} + y ^{2} \le 100 \]
Substitute the point
Why: Four squared plus 9 squared.
\[ 16 + 81 \]
Compare
Why: Ninety-seven against 100.
\[ 97 \le 100 \]
Conclude
Why: The inequality is satisfied.
Figure (svg): A coverage circle around a phone tower, with a point inside it and the exit point to the south
\[ 97 \leq 100 \;\Longrightarrow\; \text{in range} \]
Verify: compute the actual distance
Why: The square root of 97 is about 9.85 miles, just inside the 10 mile radius — with about 0.15 miles to spare. Working with the squared inequality avoided that square root entirely, which is why coverage tests are always written this way.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.3 Graph and Write Equations of Circles §9.3, pp. 628-628
Fill the middle
Example 4.
Fill in the blanks
4^2+9^2 = 97 \leq 100
Why: Sixteen plus 81 is 97, which is less than 100, so the point is inside the coverage circle. No square root was needed at any stage.
Worked example
Examples 5 and Guided Practice 6.
\[ \text{From } (4,9), \text{ how far south can you drive and stay in range? And how far west?} \]
South: x stays 4
Why: Substitute into the boundary equation.
\[ 16 + y ^{2} = 100 \]
South: solve and choose the sign
Why: Y squared is 84, and the exit is below.
\[ y\text{ about } -9.2 \]
South: find the distance
Why: From 9 down to negative 9.2.
\[ \text{about } 18.2\text{ miles} \]
West: y stays 9
Why: X squared is 19, so x is about negative 4.36.
\[ \text{about } 8.4\text{ miles} \]
Figure (svg): A coverage circle around a phone tower, with a point inside it and the exit point to the south
\[ 18.2 \text{ mi south}; \quad 8.4 \text{ mi west} \]
Verify: explain the difference between the two
Why: Driving south crosses near the middle of the circle, so the chord is long; driving west stays near the top, where the circle is narrow, so the chord is short. Both answers are chord lengths at fixed distances from the centre, and the closer to the centre a chord passes, the longer it is — with the diameter, at 20 miles, as the maximum.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.3 Graph and Write Equations of Circles §9.3, pp. 628-628
Trap
\[ 16+y^2 = 100 \;\Longrightarrow\; y = \pm\sqrt{84} \]
Take the positive value
Why: The positive root is the habitual choice.
\[ y \approx 9.2, \; \text{distance } |9-9.2| = 0.2 \quad \text{(wrong)} \]
Driving south means y decreases, so the exit point must have a negative y-coordinate. The positive root is where you would leave going north.
\[ y \approx -9.2 \;\Longrightarrow\; |9-(-9.2)| = 18.2 \]
Choose the root that matches the direction of travel
Why: The equation gives both crossings; the context picks one.
\[ \text{south from } (4,9) \text{ to } (4,-9.2) \]
Both roots are genuine points on the circle. Neither is extraneous; the context simply selects which of the two the journey reaches.
Sorting
Compare the sum of squares with 100.
Sort into buckets
Sort each location for a tower with a 10 mile radius at the origin.
The last one is only 10.05 miles out and still fails, which is why the boundary matters: coverage does not fade gradually in this model.
Comparison
Fill the blanks. Same start, different directions.
Comparison matrix
| Direction | Fixed coordinate | Miles remaining in range |
|---|---|---|
| South from (4, 9) | x = 4 | about 18.2 |
| West from (4, 9) | y = 9 | about 8.4 |
| Why they differ | the southward chord passes nearer the centre | so it is longer |
| The longest possible chord | a diameter, through the centre | 20 miles |
Every straight journey through a circle is a chord, and its length depends only on how close it passes to the centre.
Prediction
Commit before reasoning.
Predict first
Why test coverage with x squared plus y squared at most 100 rather than by computing the distance?
Correct: Because squaring avoids a square root while preserving the comparison.
\[ a,b \geq 0: \; a \leq b \;\Longleftrightarrow\; a^2 \leq b^2 \]
Why: Both quantities are non-negative, so one is smaller exactly when its square is smaller — the comparison survives squaring intact. The squared version needs two multiplications and an addition, while the direct version needs a square root as well. When a system tests thousands of points a second, that difference matters, and it is why real coverage software works entirely with squared distances.
Comparison
Fill the blanks. Each is a statement about distance.
Comparison matrix
| Curve | Defined by | Equation at the origin |
|---|---|---|
| Parabola | equal distances to a point and a line | x^2 = 4py or y^2 = 4px |
| Circle | a fixed distance from a point | x^2 + y^2 = r^2 |
| Inside a circle | distance less than the radius | x^2 + y^2 < r^2 |
| Outside a circle | distance greater than the radius | x^2 + y^2 > r^2 |
Each equation is its definition with the distance formula substituted and the square root cleared by squaring.
Pattern
One routine to graph, one to write, one to test.
The constant in the equation is r squared, never r. That single fact accounts for most of the errors in this lesson.
OpenStax Algebra and Trigonometry 2e, §2.1 The Rectangular Coordinate Systems and Graphs §2.1
Check
Rearrange, then take the root.
Check your understanding
What is the radius of the circle y^2 = -x^2 + 36?
Answer: A
Why: Rearranged, x^2 + y^2 = 36, so r^2 = 36 and r = 6.
Check
One point gives the radius.
Check your understanding
Write the equation of the circle through (2, -5) centred at the origin.
Answer: A
Why: The constant is 4 plus 25, the sum of the squared coordinates.
Check
Tangent lines. Perpendicular to the radius.
Check your understanding
What is the tangent to x^2 + y^2 = 10 at (-1, 3)?
Answer: A
Why: The radius has slope -3, so the tangent has slope 1/3.
Real world
An earthquake is felt within 40 kilometres of its epicentre. A monitoring station 24 kilometres east and 30 kilometres north of a town feels nothing.
Discussion prompt
Taking the town as the origin, decide whether an epicentre at the town itself is consistent with the station feeling nothing, and describe the region where the epicentre could be.
Hint: The station is outside the felt circle.
Answer:
\[ \text{station at } (24,30): \quad 24^2+30^2 = 576+900 = 1476 \]
\[ 40^2 = 1600, \quad \text{and } 1476 < 1600 \]
If the epicentre were at the town, the station would be about 38.4 kilometres away — inside the 40 kilometre radius, so it would have felt the quake. The town is therefore ruled out as the epicentre.
More generally the epicentre must lie more than 40 kilometres from the station, so it is somewhere outside the circle of radius 40 centred at 24 comma 30 — an inequality of exactly the form of this lesson. Real seismology does this with three or more stations at once: each one that felt the quake puts the epicentre inside a circle, each that did not puts it outside one, and the overlap of all those regions narrows the location. It is the same intersecting-circles idea as the phone-positioning problem in Lesson 9.1, run with inequalities rather than equations.
Commit first
Answer, then rate your confidence honestly.
Predict first
In the equation x squared plus y squared equals 36, is the radius 36?
Correct: No — the constant is the radius squared, so the radius is 6.
\[ 6^2+0^2 = 36 \quad \checkmark; \qquad 36^2+0^2 = 1296 \quad \times \]
Why: The equation was built by squaring the distance formula, so the right-hand side is r squared throughout. Testing settles it: if the radius were 36, the point 36 comma 0 would be on the circle, but 36 squared is 1296, not 36. Meanwhile 6 comma 0 gives exactly 36. The same convention runs through the whole chapter — the constants in conic equations are squares of the lengths they describe, which keeps radicals out of the equations at the cost of one square root when a length is actually wanted.
Explain it
They can use the distance formula and have never seen a circle's equation.
Discussion prompt
In four sentences or fewer, explain where x squared plus y squared equals r squared comes from.
Hint: Start from what a circle is.
Answer:
A circle is every point that sits the same distance from the centre. Put the centre at the origin and call that distance r.
The distance formula says the square root of x squared plus y squared equals r. Square both sides and the root disappears, leaving x squared plus y squared equals r squared — which is why the number on the right is the radius squared rather than the radius.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For rearranging, get both squared terms onto the left before reading anything. For the constant, test the point r comma 0 and see that it works. For tangents, multiply the two slopes and insist on negative 1. For roots, sketch the journey and ask whether the coordinate should be increasing or decreasing.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a circle page. Top left: draw a circle centred at the origin with a radius to a general point, label the two dashed legs, and derive the standard equation in two lines from the distance formula. Top right: graph Example 1 with all four axis points marked, and beside it write the three guided practice radii with their rearrangements shown. Middle: write the equation from the point where x is 2 and y is negative 5, showing the shortcut that avoids the square root, then check the point in your answer. Bottom left: draw Example 3's circle, the radius, and the tangent, marking the two slopes and their product. Bottom right: draw the tower's coverage circle, mark the flat tyre and both escape routes, and write one sentence explaining why the southward distance is more than twice the westward one.
If any of your radii equals the constant rather than its square root, recheck: the equation was built by squaring, so it must be unsquared to give a length.
Recap
Five things, all resting on one fixed distance.
| If you see | Then |
|---|---|
| Both variables squared with equal coefficients | A circle, once rearranged |
| A constant on the right | It is r squared, so take a root for the radius |
| A point on the circle | The sum of its squared coordinates is the constant |
| The word tangent | Take the negative reciprocal of the radius's slope |
| A coverage region | Use an inequality, not an equation |
| A journey leaving a region | Fix one coordinate, solve, and pick the matching root |
Lesson 9.4 stretches the circle into an ellipse, where the two squared terms get different denominators and the single radius becomes two different axis lengths.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.3 Graph and Write Equations of Circles §9.3, pp. 626-633 — everything on these slides traces back here
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