Appendix A — Answer Key, Chapter 15: Equations of Circles

SOL G.PC.4 (a, b) · Covers textbook Chapter 15 and the companion workbook. Item numbers match the textbook; workbook items are the same problems, so this key serves both. Item numbers run continuously from 1 to 112 across the chapter.

Conventions used in every answer below. Standard form is $(x-h)^2+(y-k)^2=r^2$, with center $(h, k)$ and radius $r$. Both $x$- and $y$-terms are subtractions, so a negative center coordinate flips the sign printed in the equation. The right-hand side is $r^2$ — reading a radius off an equation always costs one square root. Radii are given in simplest radical form when $r^2$ is not a perfect square.

Every one of the four ends at the same two computations — a center and a radius. Before working a problem, decide which one you still have to find.

Lesson 15.1 — Deriving the Equation

Guided practice

1. The horizontal leg, $x-2$, and the vertical leg, $y-3$.
2. The hypotenuse is $r$, and the Pythagorean Theorem applied to the triangle gives $(x-2)^2+(y-3)^2=r^2$.
3. $x^2+y^2=36$, with $h=0$ and $k=0$.
4. Both subtractions, $x-h$ and $y-k$, vanish when $h=0$ and $k=0$, leaving nothing to write.
5. Center $(5, 2)$, radius $7$.
6. Write $(x-5)^2+(y-2)^2=r^2$; substitute $r=7$ to get $(x-5)^2+(y-2)^2=7^2$; square the $7$ to get $(x-5)^2+(y-2)^2=49$.

Independent practice

7. $x^2+y^2=$ $9$
8. $x^2+y^2=$ $64$
9. $(x-4)^2+y^2=$ $25$
10. $x^2+(y+6)^2=$ $36$
11. $(x-5)^2+(y-2)^2=$ $49$
12. $(x+3)^2+(y-1)^2=$ $16$
13. $(x+2)^2+(y+5)^2=$ $81$
14. $(x-6)^2+(y+4)^2=$ $100$
15. $3^2+4^2=25$. Yes.
16. $2^2+2^2=8 \ne 9$. No.
17. $(7-3)^2+(1-1)^2=16$. Yes.
18. $(5-1)^2+(5-2)^2=16+9=25$. Yes.
19. $(0-3)^2+(0-4)^2=9+16=25$. Yes.
20. $(1+2)^2+(1-3)^2=9+4=13 \ne 20$. No.
21. legs $6$ and $8$; $r^2=100$. $x^2+y^2=100$
22. legs $3$ and $4$; $r^2=25$. $(x-1)^2+(y-1)^2=25$
23. legs $3$ and $-4$; $r^2=25$. $(x+2)^2+(y-3)^2=25$
24. legs $9$ and $12$; $r^2=81+144=225$. $x^2+y^2=225$
25. The student wrote both terms as if the center's own sign had been copied in directly. With center $(2, -3)$: $h=2$ gives $(x-2)$, not $(x+2)$; $k=-3$ gives $(y+3)$, not $(y-3)$. Correct: $(x-2)^2+(y+3)^2=25$.
26. The right side is $r^2$ because the Pythagorean Theorem itself produces $\text{leg}^2+\text{leg}^2=\text{hypotenuse}^2$ — squares of the legs summing to the square of the radius. Nothing in the derivation ever isolates $r$ by itself; that square root is a separate step taken only when a radius, not $r^2$, is what's wanted.

Exit ticket 15.1

27. $x^2+(y-5)^2=$ $9$
28. $4^2+3^2=16+9=25$. Yes, by the Pythagorean Theorem: the legs from the origin to $(4,3)$ square-sum to $25$, matching $r^2$.

Lesson 15.2 — Reading the Equation and the Graph

Guided practice

29. Center $(-3, 4)$, radius $6$.
30. The term is $(x-h)$ with $h=-3$, which is $x-(-3)=x+3$ — subtracting a negative number turns the sign to a plus.
31. Center $(2, -3)$, radius $7$.
32. Straight up, down, left, or right from the center — those are the directions where the circle crosses a grid line exactly, so the distance can be counted rather than computed.
33. $(x-3)^2+(y+4)^2=18$; center $(3, -4)$; radius $3\sqrt{2}$.
34. Legs $2$ and $2$; $r^2=2^2+2^2=8$; radius $2\sqrt{2}$.

Independent practice

35. center $(4, 1)$, radius $6$
36. center $(-2, 5)$, radius $7$
37. center $(6, -3)$, radius $8$
38. center $(0, 7)$, radius $9$
39. center $(-5, 0)$, radius $10$
40. center $(1, -8)$, radius $12$
41. center $(-9, -2)$, radius $13$
42. center $(0, 0)$, radius $11$
43. center $(2, 1)$, radius $3\sqrt{2}$
44. center $(-1, 4)$, radius $2\sqrt{5}$
45. center $(0, -3)$, radius $5\sqrt{2}$
46. center $(3, 0)$, radius $3\sqrt{5}$
47. $r=4-(-3)=7$. $(x-2)^2+(y+3)^2=49$
48. $r=1-(-5)=6$. $(x+5)^2+(y-1)^2=36$
49. $r=0-(-9)=9$. $x^2+y^2=81$
50. $r^2=(2\sqrt3)^2=12$. $(x-7)^2+(y+5)^2=12$
51. $r^2=15^2=225$. $(x+6)^2+(y+6)^2=225$
52. $6-(-2)=8$, confirming $r=8$. $(x-5)^2+(y+2)^2=64$
53. They read the center's $y$-coordinate as $2$ instead of $-2$ — the term is $(y-k)$ with $y+2$ meaning $k=-2$, not $2$. And they gave $r=49$ instead of taking the square root: $r=\sqrt{49}=7$. Correct: center $(4, -2)$, radius $7$.
54. The right side of the equation is $r^2$, produced by squaring the radius during the derivation — recovering $r$ means undoing that square, which is a square root. The numbers subtracted from $x$ and $y$ are $h$ and $k$ themselves, already in the form the derivation wrote them in — no undoing is needed, only reading the sign correctly.

Exit ticket 15.2

55. center $(-3, 6)$, radius $2\sqrt{10}$
56. $r=4-(-2)=6$. $(x-1)^2+(y-4)^2=36$

Lesson 15.3 — Writing the Equation from Two Points

Guided practice

57. Legs $3$ and $4$; radius $5$.
58. $(x-1)^2+(y-2)^2=25$
59. Center $(2, 2)$, found as the midpoint of $A(-2,-1)$ and $B(6,5)$: $\left(\dfrac{-2+6}{2}, \dfrac{-1+5}{2}\right)=(2,2)$.
60. Radius $5$, measured from the center $(2,2)$ to endpoint $B(6,5)$ — not the segment $\overline{AB}$ itself.
61. Using $\overline{AB}$'s full length as the radius would make the circle's radius (and therefore its area) far too large — $\overline{AB}$ is the diameter, twice the true radius.
62. Both methods end at a center and a radius. The center and a point method is already given the center and has to find the radius. The diameter endpoints method has to find both: the center by averaging, then the radius by measuring from that new center.

Independent practice

63. legs $4$ and $3$; $r^2=25$. $(x-3)^2+(y-5)^2=25$
64. legs $12$ and $-5$; $r^2=169$. $(x+2)^2+(y-4)^2=169$
65. legs $-8$ and $6$; $r^2=100$. $x^2+y^2=100$
66. legs $0$ and $8$; $r^2=64$. $(x-1)^2+(y+3)^2=64$
67. legs $-6$ and $8$; $r^2=100$. $(x-4)^2+(y+1)^2=100$
68. legs $8$ and $-6$; $r^2=100$. $(x+5)^2+(y-2)^2=100$
69. legs $5$ and $12$; $r^2=169$. $x^2+(y-3)^2=169$
70. legs $-9$ and $-12$; $r^2=225$. $(x-6)^2+(y-6)^2=225$
71. midpoint $(3,4)$; $r^2=25$. $(x-3)^2+(y-4)^2=25$
72. midpoint $(0,4)$; $r^2=25$. $x^2+(y-4)^2=25$
73. midpoint $(2,2)$; $r^2=49$. $(x-2)^2+(y-2)^2=49$
74. midpoint $(-2,1)$; $r^2=25$. $(x+2)^2+(y-1)^2=25$
75. midpoint $(5,6)$; $r^2=25$. $(x-5)^2+(y-6)^2=25$
76. midpoint $(1,2)$; $r^2=41$. $(x-1)^2+(y-2)^2=41$
77. midpoint $(-2,0)$; $r^2=52$. $(x+2)^2+y^2=52$
78. midpoint $(2,-6)$; $r^2=64$. $(x-2)^2+(y+6)^2=64$
79. midpoint $(3,2)$; $r^2=25$. $(x-3)^2+(y-2)^2=25$
80. legs $0$ and $12$; $r^2=144$. $(x-4)^2+(y+3)^2=144$
81. The correct midpoint is $\left(\dfrac{2+10}{2}, \dfrac{6+2}{2}\right)=$ $(6, 4)$, not the sum of the coordinates. Distance from $(6,4)$ to $(10,2)$: legs $4$ and $-2$, $r^2=20$. Correct equation: $(x-6)^2+(y-4)^2=20$.
82. Center and a point is already given the center, so it skips straight to computing the radius by distance. Diameter endpoints is given neither directly — it computes the center first, by midpoint, and only then can compute the radius the same way the other method does.

Exit ticket 15.3

83. legs $4$ and $3$; $r^2=25$. $(x-5)^2+(y-5)^2=25$
84. midpoint $(4,3)$; $r^2=25$. $(x-4)^2+(y-3)^2=25$

Lesson 15.4 — Putting It Together

Guided practice

85. Equation — read $h$, $k$; take $\sqrt{\ }$ of the right side. Graph — read the center; measure to the edge. Diameter endpoints — center = midpoint; $r$ = distance to an endpoint. Center and a point — $r$ = distance from center to the point.
86. Wrong: $(x-3)^2+(y-4)^2=36$. Correct: $(x+3)^2+(y-4)^2=36$. Only the sign on the $x$-term changed — the $y$-term and the right side were already correct.
87. $x^2+y^2=9$ and $x^2+y^2=36$. Only the right-hand side ($r^2$) changed; both centers stayed at the origin.
88. $(x+4)^2+y^2=16$ and $(x-4)^2+y^2=16$. Only $h$ changed, from $-4$ to $4$; the radius and the $y$-term stayed the same.
89. Cell tower — center and a point. Garden bed — diameter endpoints. Radar — a graph.
90. Answers vary by panel; each should give the drawn center's coordinates and count grid units to the circle for the radius.

Independent practice

91. center $(5, -2)$, radius $9$
92. center $(-7, 3)$, radius $12$
93. center $(0, -9)$, radius $10$
94. center $(8, 8)$, radius $13$
95. $r=3-(-5)=8$. $(x-3)^2+(y+5)^2=64$
96. $r=0-(-6)=6$. $(x+6)^2+y^2=36$
97. midpoint $(4,1)$; $r^2=25$. $(x-4)^2+(y-1)^2=25$
98. midpoint $(-1,-2)$; $r^2=25$. $(x+1)^2+(y+2)^2=25$
99. legs $8$ and $6$; $r^2=100$. $(x-2)^2+(y+6)^2=100$
100. legs $0$ and $-14$; $r^2=196$. $(x+4)^2+(y-4)^2=196$
101. midpoint $(-4,1)$; $r^2=25$. $(x+4)^2+(y-1)^2=25$
102. legs $3$ and $-3$; $r^2=18$. $x^2+y^2=18$, radius $3\sqrt2$
103. center $(4, -1)$, radius $5\sqrt2$
104. center $(-2, -3)$, radius $6\sqrt2$
105. legs $8$ and $15$; $r^2=289$. $x^2+y^2=289$
106. midpoint $(-4,0)$; $r^2=25$. $(x+4)^2+y^2=25$
107. legs $0$ and $8$; $r^2=64$. $(x-3)^2+(y+6)^2=64$
108. They gave $r^2$ as the radius. The radius is $\sqrt{40}=$ $2\sqrt{10}$.
109. A midpoint is the point exactly between two others — an average of coordinates. A segment's length is a distance, found with the Pythagorean Theorem. Neither computation can substitute for the other, and the diameter-endpoints method genuinely needs both in sequence: the midpoint to locate the center, then a length from that new point to find the radius.
110. $c$ must be positive. $c=r^2$, and a squared real number is never negative — so $c>0$ gives a genuine circle of radius $\sqrt{c}$, $c=0$ gives a single point (every "leg" is forced to $0$), and $c<0$ describes no real points at all, since no real $x$ and $y$ can make two squares sum to a negative number.

Exit ticket 15.4

111. legs $6$ and $-8$; $r^2=100$. $(x+3)^2+(y-5)^2=100$
112. "Is a center and a radius already given, or does one — or both — still need to be found, and from what?"

Chapter 15 Review — answers

Review 1 (G.PC.4a).

• horizontal leg $2-(-4)=$ $6$; vertical leg $10-2=$ $8$
• $r^2=6^2+8^2=$ $100$, so $r=$ $10$
• $(x+4)^2+(y-2)^2=100$
• Distance from $C(-4,2)$ to $R(-10,-4)$: legs $-6$ and $-6$, so $r^2=36+36=72 \ne 100$. $R$ is not on the circle — its squared distance from the center doesn't match $r^2$, so there's no need to write out a new equation to know it fails.

Review 2 (G.PC.4b).

• center $(6, -9)$, radius $13$
• $r=5-(-3)=8$. $(x+4)^2+(y-5)^2=64$
• The first circle's radius, $13$, is bigger by $5$ than the second circle's radius, $8$.
• The right side of an equation is $r^2$, so undoing the square to reach $r$ needs a square root. The center's coordinates, $h$ and $k$, are already the actual numbers subtracted in the equation — reading them back out is just reading, with the sign flipped.

Review 3 (G.PC.4b).

• Diameter $A(-7,-4)$, $B(1,2)$: midpoint $(-3,-1)$; $r^2=25$. $(x+3)^2+(y+1)^2=25$
• Center $(-3,-8)$, point $(9,-3)$: legs $12$ and $5$; $r^2=169$. $(x+3)^2+(y+8)^2=169$
• First circle: diameter endpoints. Second circle: center and a point.
• Both methods still had to compute a radius — the first from a midpoint it also had to find first, the second directly from the given center — but neither one could skip that final distance calculation.

Every item in Chapter 15 is answered above: 1 to 112, plus the three chapter reviews.