Question

How do you write an equation for a circle with center (6, –8) that passes through (0, 0)?

Answer 1

`x^2+y^2-12x+16y=0`.

Explanation:

Centre of the circle is `C(6,-8)` and since the circle passes through

the Origin `O(0,0)`,

the radius `r` is given by, `r"=dist."CP`

`:. r^2=(6-0)^2+(-8-0)^2=36+64=100`

`:." the eqn. is, "(x-6)^2+(y+8)^2=100`, or,

`x^2+y^2-12x+16y=0`.