Question

How do you find the equation for a circle given center at (-1,-2) & passes through point (15,-2)?

Answer 1

`(x+1)^2 +(y+2)^2 = 256`

Explanation:

The standard form of a circle is `(x-h)^2 + (y-k)^2 = r^2` where `(h,k)` is the centre and `r` is the radius.

This circle is therefore `(x+1)^2 +(y+2)^2 = r^2`

We can use the given point to find `r` but substitution.

`(15+1)^2 +(-2+2)^2 = r^2`

`r^2 = 16^2 = 0`
`:. r= 16`

The required equation is `(x+1)^2 +(y+2)^2 = 256`