Question

How do you find the equation of a circle in standard form given C(2,-4) and r=8?

Answer 1

Equation of a circle in standard form is `x^2+y^2-4x+8y-44=0`

Explanation:

If the center is `C(2,-4)` and radius is `8`,

the point `(x,y)` on the circle moves so that its distance from `C(2,-4)` is always `8`

As distance of point `(x,y)` from `C(2,-4)` is `8`, we have

`sqrt((x-2)^2+(y-(-4))^2)=8`

or `(x-2)^2+(y+4)^2=8^2`

or `x^2-4x+4+y^2+8y+16=64`

or `x^2+y^2-4x+8y-64+4+16=0`

or `x^2+y^2-4x+8y-44=0`
graph{x^2+y^2-4x+8y-44=0 [-19, 21, -13.68, 6.32]}