Question

How do you find the center and radius for `x^2 - 3x + y^2 - 3y - 20 = 0`?

Answer 1

The center is `=(3/2,3/2)` and the radius `r=7sqrt2/2`

Explanation:

The standard equation of a circle is

`(x-a)^2+(y-b)^2=r^2`

where

The center is `(a,b)` and the radius is `=r`

Here, we have

`x^2-3x+y^2-3y-20=0`

Complete the square for `x` and `y`

`x^2-3x+9/4+y^2-3y+9/4=20+9/4+9/4`

`(x-3/2)^2+(y-3/2)^2=98/4=(7sqrt2/2)^2`

The center is `=(3/2,3/2)` and the radius `r=7sqrt2/2`

graph{x^2-3x+y^2-3y-20=0 [-16.35, 15.68, -4.36, 11.66]}