Question

Find equation of circle with center on a line y= - x has radius 4 and passes through origin?

Answer 1

Equation of circle is `x^2+y^2+4sqrt2x-4sqrt2y=0`

Explanation:

As circle passes thrugh origin, its cinstant term ought to be `0` and hence equation is of the type

`x^2+y^2+2gx+2fy=0`, whose center is `(-g,-f)` and radius is `sqrt(g^2+f^2)`

As center `(-g,-f)` lies on `y=-x`,

we have `-f=-(-g)` or `f=-g`

and aas radius is `4`, we have `sqrt(g^2+f^2)=4`

or `2g^2=16` i.e. `g=sqrt8=2sqrt2` and `f=-2sqrt2`

and equation of circle is `x^2+y^2+4sqrt2x-4sqrt2y=0`

graph{(x^2+y^2+4sqrt2x-4sqrt2y)(y+x)=0 [-12.79, 7.21, -2.92, 7.08]}