Equation of a Circle: Geometry Math Worksheet Answer Key – PDF Download

Geometry Equation Circle Worksheet Answer Key: problems 1–5

Problem 1: Center at the origin

With the center at the origin, \(x^{2}+y^{2}=r^{2}\) and \(r^{2}=25\).

\(x^{2}+y^{2}=25\)

Problem 2: Equation from center and radius

Use \((x-h)^{2}+(y-k)^{2}=r^{2}\) with \(h=-5\) and \(k=-2\). Subtracting a negative number adds: \(x-(-5)=x+5\) and \(y-(-2)=y+2\). Also \(r^{2}=(\sqrt{11})^{2}=11\).

\((x+5)^{2}+(y+2)^{2}=11\)

Problem 3: Negative x-coordinate

Here \(h=-2\) and \(k=5\), so \(x-h=x+2\) and \(y-k=y-5\). Also \(r^{2}=9\).

\((x+2)^{2}+(y-5)^{2}=9\)

Problem 4: Radius is a square root

\(r^{2}=(\sqrt{7})^{2}=7\).

\((x-4)^{2}+(y+1)^{2}=7\)

Problem 5: Given the diameter length

The radius is half the diameter: \(r=10\), so \(r^{2}=100\).

\((x+6)^{2}+(y+3)^{2}=100\)

Geometry Equation Circle Worksheet Answer Key: problems 6–10

Problem 6: Read the center and radius

Compare with \((x-h)^{2}+(y-k)^{2}=r^{2}\): \(x-5\) gives \(h=5\); \(y+2=y-(-2)\) gives \(k=-2\); \(r^{2}=49\) gives \(r=7\).

Center \((5,-2)\), radius 7

Problem 7: Circle at the origin

There are no shifts, so the center is the origin. \(r^{2}=36\) gives \(r=6\).

Center \((0,0)\), radius 6

Problem 8: Radius that is not a whole number

\(x+1=x-(-1)\) gives \(h=-1\), and \(k=4\). The radius is \(r=\sqrt{20}=\sqrt{4\cdot 5}=2\sqrt{5}\).

Center \((-1,4)\), radius \(2\sqrt{5}\)

Problem 9: Unit circle shifted

Both signs flip: \(h=-8\) and \(k=-8\). \(r^{2}=1\) gives \(r=1\).

Center \((-8,-8)\), radius 1

Problem 10: Is the point on the circle?

Substitute: \((6-2)^{2}+(6-3)^{2}=4^{2}+3^{2}=16+9=25\).

This equals \(r^{2}=25\), so the point lies on the circle.

Geometry Equation Circle Worksheet Answer Key: problems 11–15

Problem 11: Inside, on or outside

A: \((4-1)^{2}+(2+2)^{2}=9+16=25\), and 25 is less than 36, so A is inside.

B: \((7-1)^{2}+(-2+2)^{2}=36+0=36\), which equals 36, so B is on the circle.

C: \((6-1)^{2}+(5+2)^{2}=25+49=74\), and 74 is greater than 36, so C is outside.

Problem 12: Center and a point on the circle

The radius is the distance from the center to the point: \(r^{2}=(4-1)^{2}+(6-2)^{2}=9+16=25\).

\((x-1)^{2}+(y-2)^{2}=25\)

Problem 13: Endpoints of a diameter

The center is the midpoint: \((\frac{-2+4}{2},\frac{1+9}{2})=(1,5)\).

The radius is the distance from the center to B: \(r^{2}=(4-1)^{2}+(9-5)^{2}=9+16=25\).

\((x-1)^{2}+(y-5)^{2}=25\)

Problem 14: Find a missing coordinate

The equation is \((x-2)^{2}+(y+3)^{2}=25\). Substitute \(x=5\): \(9+(k+3)^{2}=25\).

So \((k+3)^{2}=16\) and \(k+3=4\) or \(k+3=-4\).

\(k=1\) or \(k=-7\)

Problem 15: Tangent to the x-axis

A circle tangent to the x-axis touches it where the radius is perpendicular to the axis, so the radius equals the distance from the center to the axis: the y-coordinate, \(r=4\).

\((x-3)^{2}+(y-4)^{2}=16\)

Geometry Equation Circle Worksheet Answer Key: problems 16–20

Problem 16: Tangent to the y-axis

The distance from the center to the y-axis is the absolute value of the x-coordinate: \(r=|-5|=5\), so \(r^{2}=25\).

\((x+5)^{2}+(y-2)^{2}=25\)

Problem 17: Complete the square, case 1

Group the terms and move the constant: \((x^{2}+8x)+(y^{2}-2y)=-8\).

Add \((\frac{8}{2})^{2}=16\) for the x-terms and \((\frac{-2}{2})^{2}=1\) for the y-terms to both sides: \((x^{2}+8x+16)+(y^{2}-2y+1)=-8+16+1=9\).

Factor each perfect square: \((x+4)^{2}+(y-1)^{2}=9\).

Center \((-4,1)\), radius \(3\)

Problem 18: Complete the square, case 2

Group the terms and move the constant: \((x^{2}-10x)+(y^{2}-4y)=-4\).

Add \((\frac{-10}{2})^{2}=25\) for the x-terms and \((\frac{-4}{2})^{2}=4\) for the y-terms to both sides: \((x^{2}-10x+25)+(y^{2}-4y+4)=-4+25+4=25\).

Factor each perfect square: \((x-5)^{2}+(y-2)^{2}=25\).

Center \((5,2)\), radius \(5\)

Problem 19: Complete the square, case 3

Group the terms and move the constant: \((x^{2}+2x)+(y^{2}+6y)=15\).

Add \((\frac{2}{2})^{2}=1\) for the x-terms and \((\frac{6}{2})^{2}=9\) for the y-terms to both sides: \((x^{2}+2x+1)+(y^{2}+6y+9)=15+1+9=25\).

Factor each perfect square: \((x+1)^{2}+(y+3)^{2}=25\).

Center \((-1,-3)\), radius \(5\)

Problem 20: Only one variable shifted

Group the terms and move the constant: \(x^{2}+(y^{2}-12y)=-20\).

Add \((\frac{-12}{2})^{2}=36\) for the y-terms to both sides: \(x^{2}+(y^{2}-12y+36)=-20+36=16\).

Factor each perfect square: \(x^{2}+(y-6)^{2}=16\).

Center \((0,6)\), radius \(4\)

Geometry Equation Circle Worksheet Answer Key: problems 21–25

Problem 21: Points with a given x-coordinate

Substitute \(x=8\): \((8-2)^{2}+(y+1)^{2}=100\), so \(36+(y+1)^{2}=100\).

\((y+1)^{2}=64\), so \(y+1=8\) or \(y+1=-8\).

\(y=7\) or \(y=-9\)

Problem 22: Cell tower coverage

Edge of coverage: \((x-3)^{2}+(y+4)^{2}=100\).

Town A: \((9-3)^{2}+(4+4)^{2}=36+64=100\), exactly on the edge.

Town B: \((-3-3)^{2}+(2+4)^{2}=36+36=72\), less than 100, so inside.

Town C: \((12-3)^{2}+(-10+4)^{2}=81+36=117\), more than 100, so outside.

A is on the boundary (signal at the limit), B has a signal, C has none.

Problem 23: Wi-Fi router

Boundary: \((x-2)^{2}+(y-1)^{2}=625\), since \(25^{2}=625\).

Laptop at (20, 18): \((20-2)^{2}+(18-1)^{2}=324+289=613\), less than 625, so inside the range.

Laptop at (20, 20): \((20-2)^{2}+(20-1)^{2}=324+361=685\), more than 625, so outside the range.

Problem 24: Pond on a map

Group and complete the square: \((x^{2}-8x+16)+(y^{2}+6y+9)=16+9\), so \((x-4)^{2}+(y+3)^{2}=25\).

Center \((4,-3)\), radius 5 yd.

Origin: \((0-4)^{2}+(0+3)^{2}=16+9=25\), which equals 25, so the entrance is on the edge of the pond.

Problem 25: Earthquake zone

Boundary: \((x+6)^{2}+(y-8)^{2}=225\).

Point (6, −1): \((6+6)^{2}+(-1-8)^{2}=144+81=225\), exactly on the boundary, so it is felt right at the limit.

Origin: \((0+6)^{2}+(0-8)^{2}=36+64=100\), less than 225, so felt.

How to use this geometry equation circle worksheet answer key

Read this geometry equation circle worksheet answer key after trying each problem yourself, and redo any problem you missed.

See every chapter on the Geometry math page.