Geometry Equations of Circles and Parabolas Worksheet Solutions (PDF)
This circle equations answers for Geometry covers equations of circles and parabolas (Common Core G-GPE.A.1, G-GPE.A.2), with a free PDF to print.
Geometry · Common Core State Standards: G-GPE.A.1, G-GPE.A.2
Circle equations answers: exercises 1 to 10
Exercise 1 – Circle at the origin
\(x^{2} + y^{2} = 7^{2}\)
x² + y² = 49
Exercise 2 – Center and radius given
\((x – 3)^{2} + (y – (-2))^{2} = 5^{2}\)
(x − 3)² + (y + 2)² = 25
Exercise 3 – Read the equation
The center has opposite signs: \((-4, 1)\). The radius is \(\sqrt{36} = 6\).
Center (−4, 1); radius 6
Exercise 4 – Radius from r squared
\(r = \sqrt{20} = 2\sqrt{5}\)
2√5
Exercise 5 – Point on the circle?
\((4 – 1)^{2} + (6 – 2)^{2} = 9 + 16 = 25\).
Yes
Exercise 6 – Outside or inside?
B: \(6^{2} + 1^{2} = 37 > 25\), outside.
D: \(1^{2} + 1^{2} = 2 < 25\), inside.
B is outside; D is inside
Exercise 7 – Circle through a point
\(r^{2} = (3 + 1)^{2} + (5 – 2)^{2} = 16 + 9 = 25\).
(x + 1)² + (y − 2)² = 25
Exercise 8 – Circle from a diameter
Center = midpoint: \((2, 5)\).
Diameter: \(d^{2} = 8^{2} + 4^{2} = 80\), so \(r^{2} = \frac{80}{4} = 20\).
(x − 2)² + (y − 5)² = 20
Exercise 9 – Complete the square (1)
\((x^{2} – 6x + 9) + (y^{2} + 4y + 4) = 12 + 9 + 4\)
\((x – 3)^{2} + (y + 2)^{2} = 25\)
Center (3, −2); radius 5
Exercise 10 – Complete the square (2)
\((x^{2} – 12x + 36) + (y^{2} + 6y + 9) = -20 + 36 + 9\)
\((x – 6)^{2} + (y + 3)^{2} = 25\)
Center (6, −3); radius 5
Exercises 11 and up
Exercise 11 – Complete the square (3)
\((x^{2} – 10x + 25) + y^{2} = 25\)
\((x – 5)^{2} + y^{2} = 25\)
Center (5, 0); radius 5
Exercise 12 – Tangent to the x-axis
The radius equals the distance to the x-axis: \(r = 3\).
\((x – 4)^{2} + (y – 3)^{2} = 9\)
The point of tangency is directly below the center: \((4, 0)\).
(x − 4)² + (y − 3)² = 9; (4, 0)
Exercise 13 – Line and circle
Substitute: \(x^{2} + (x + 1)^{2} = 25\).
\(2x^{2} + 2x – 24 = 0\), so \(x^{2} + x – 12 = 0\) and \((x + 4)(x – 3) = 0\).
\(x = 3\) gives \((3, 4)\); \(x = -4\) gives \((-4, -3)\).
(3, 4) and (−4, −3)
Exercise 14 – Area and circumference
\(r = 9\).
\(C = 2\pi \cdot 9 = 18\pi\) and \(A = \pi \cdot 9^{2} = 81\pi\).
18π and 81π
Exercise 15 – Parabola equation
A point \((x, y)\) is equally far from the focus and the directrix.
\(\sqrt{x^{2} + (y – 2)^{2}} = y + 2\)
\(x^{2} + y^{2} – 4y + 4 = y^{2} + 4y + 4\)
\(x^{2} = 8y\)
x² = 8y
Exercise 16 – Parabola with focus (0, 3)
Use \(x^{2} = 4py\) with \(p = 3\).
x² = 12y
Exercise 17 – Check a point on a parabola
\(4^{2} = 16 = 8 \cdot 2\), so P is on the parabola.
Distance to the focus: \(4\). Distance to the directrix: \(2 – (-2) = 4\). They are equal.
Both distances are 4
Exercise 18 – Find the focus
Write \(x^{2} = 16y\), so \(4p = 16\) and \(p = 4\).
Focus (0, 4); directrix y = −4
Exercise 19 – Radio coverage
\((x – 10)^{2} + (y + 5)^{2} = 169\).
City: \((22 – 10)^{2} + (0 + 5)^{2} = 144 + 25 = 169\).
(x − 10)² + (y + 5)² = 169; the city is exactly on the boundary
Exercise 20 – Satellite dish
\(x^{2} = 24y\), so \(4p = 24\) and \(p = 6\): the focus is \((0, 6)\), 6 feet above the vertex.
At width 24, \(x = 12\) and \(y = \frac{144}{24} = 6\).
Receiver at (0, 6); depth 6 ft
Key facts behind this circle equations answers
- Circle: (x – h)^2 + (y – k)^2 = r^2, center (h, k), radius r.
- Read the center with opposite signs from the equation.
- Inside, on, or outside: compare (x – h)^2 + (y – k)^2 with r^2.
- Complete the square to convert x^2 + y^2 + Dx + Ey + F = 0.
- Parabola with focus (0, p) and directrix y = -p: x^2 = 4py.
How to use this circle equations answers
Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.
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Download the circle equations answers as a PDF with the button below and keep it with your notes.
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