Question

How do you write the equation of the circle with the given center at (-2,4) and passes through the point at (1,-7)?

Answer 1

Equation of the circle is `(x+2)^2+(y-4)^2=121` or `x^2+y^2+4x-8y-101=0`

Explanation:

Equation of a circle with center at `(h,k)` is

`(x-h)^2+(y-k)^2=r^2`, where `r` is its radius.

Hence equation of a circle with center at `(-2,4)` is

`(x-(-2))^2+(y-4)^2=r^2` or `(x+2)^2+(y-4)^2=r^2`

As it passes through `(1,-7)`, we will have

`(-2+2)^2+(-7-4)^2=r^2`

or `0^2+11^2=r^2` and `r^2=0+121=121`

Hence, equation of circle is `(x+2)^2+(y-4)^2=121`

or `x^2+y^2+4x-8y-101=0` and its radius is `sqrt121=11`
graph{x^2+y^2+4x-8y-101=0 [-28, 24, -10, 16]}