A circle's equation is the distance formula in disguise. Every point (x,y) on a circle with center (h,k) and radius r sits exactly r units from the center, so the Pythagorean theorem gives the standard form (x−h)2+(y−k)2=r2.
Reading the center takes care with signs. The equation subtracts h and k, so (x+3)2 means h=−3. The right side is r2, not r, so take a square root to get the radius.
If the equation is expanded, like x2+y2+8x−2y−8=0, complete the square to get back to standard form.
Worked example. Find the center and radius of x2+y2+8x−2y−8=0. Group the terms and move the constant to get (x2+8x)+(y2−2y)=8. Half of 8 is 4, and 42=16. Half of −2 is −1, and (−1)2=1. Add 16 and 1 to both sides to get (x+4)2+(y−1)2=25. The center is (−4,1) and the radius is 5.
To test whether a point is on the circle, plug it in. The point (−1,5) gives 32+42=25, so it lies on the circle.
- Standard form: The equation (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2(x−h)2+(y−k)2=r2, where (h,k)(h, k)(h,k) is the center and rrr is the radius.
- General form: The expanded version of a circle equation, such as x2+y2+Dx+Ey+F=0x^2 + y^2 + Dx + Ey + F = 0x2+y2+Dx+Ey+F=0, which hides the center and radius.
- Completing the square: Adding the square of half the linear coefficient to both sides so each group factors into a perfect square.
Check yourself. What are the center and radius of (x−2)2+(y+5)2=49? (Answer: The center is (2,−5) and the radius is 7.)