Each solution shows the substitution or the completing-the-square steps, then the final answer in bold. Check that the signs in the parentheses are opposite to the center coordinates and that the right side is the square of the radius.
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: Center and radius given
Use \((x-h)^{2}+(y-k)^{2}=r^{2}\) with \(h=3\), \(k=2\) and \(r^{2}=16\). Since \(r^{2}=4^{2}=16\):
\((x-3)^{2}+(y-2)^{2}=16\)
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 \(r^{2}=100\).
\((x+6)^{2}+(y+3)^{2}=100\)
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.
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, \(r=4\).
\((x-3)^{2}+(y-4)^{2}=16\)
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 \(4\)
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.
See every chapter on the Geometry math page.