Question

How do you find the center and radius of the circle with this equation `x^2 + y^2 – 8x – 20y + 80 = 0`?

Answer 1

The centre is `(4,10)` and the radius is `sqrt98`.

Explanation:

You would have to complete the square twice, once for x and once for y, in order to write in in standard form `(x-a)^2+(y-b)^2=r^2`, from which the centre would be `(a,b)` and the radius `r`.

Completing the square to write it in this form yields :

`x^2-8x+((-8)/2)^2-((-8)/2)^2+y^2-20y+((-20)/2)^2-((-20)/2)^2+80=0`

`therefore(x-4)^2+(y-10)^2-16-100+80=0`

`therefore(x-4)^2+(y-10)^2=98`.

Hence the centre is `(4,10)` and the radius is `sqrt98`.