Question

If the circles x^2+y^2+2ax+c^2=0 and x^2+y^2+2by+c^2=0 touch each other,prove that 1/a^2+1/b^2=1/c^2?

`1/a^2+1/b^2=1/c^2`

Answer 1

See below

Explanation:

If the two circles touch each other, they must have one and only one point in common.

This point must satisfy both

`x^2+y^2+2ax+c^2=0`
and
`x^2+y^2+2by+c^2=0`

so that `ax=by`

Substituting `y=a/b x` in the first equation gives us a quadratic for `x`,

`(1+a^2/b^2)x^2+2ax+c^2=0`

Since the circles touch each other, this quadratic must have equal roots, so

`(2a)^2=4(1+a2/b^2)c^2 implies 1/c^2=1/a^2(1+a^2/b^2) =1/a^2+1/b^2`