Question

How do you find the equation for a circle that passes through (2,-3) with center (1,3)?

Answer 1

`(x-1)^2+(y-3)^2=13`

Explanation:

Any circle with centre `(a,b)` and radius `r=sqrt(x^2+y^2)` has equation `(x-a)^2+(y-b)^2=r^2`.

So in this case, `r=sqrt(2^2+3^2)=sqrt13`.

So equation is `(x-1)^2+(y-3)^2=13`.