Question

Find the equation of circle which is concentric to `x^2+y^2-8x+4=0` and touches line `x+2y+6=0`?

Answer 1

`x^2+y^2-8x-4=0`

Explanation:

The center of circle `x^2+y^2-8x+4=0` is `(4,0)`

As the desired circle is concentric to this circle, its center too is `(4,0)`

and radius is the length of perpendicular from `(4,0)` to `x+2y+6=0`

i.e. `|(4+2xx0+6)/(sqrt(1^2+2^2))|=10/sqrt5`

Hence equation of circle is

`(x-4)^2+y^2=100/5=20`

or `x^2-8x+16+y^2-20=0`

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

graph{(x+2y+6)(x^2+y^2-8x-4)(x^2+y^2-8x+4)=0 [-10, 10, -5, 5]}