I'm trying to find the locus of all points so that their quotient of the Tangents lengths to circles: $x^2+y^2-12x=0, x^2+y^2+8x-3y=0$ is $2:3$, respectively.
i tried to use the formula: $(-ma+b-n)^2=R^2(m^2+1)$, but it didn't help. any ideas? thank you.
Hints: Circles
$$C_1:\;\;(x-6)^2+y^2=6^2\;\;\;\;,\;\;\;\;\;C_2:\;\;(x+4)^2+\left(y-\frac{3}{2}\right)^2=\frac{73}{4}$$
Since a tangent line to a circle is perpendicular to the circle's radius to the tangency point and since Pythagoras was a great guy, a tangent segment's length from an exterior point of a circle to the tangency point is given by the square root of the difference of the squares of the lengths of the segment between the point to the circle's center and its radius, so if $\,(a,b)\,$ is one of the points wanted then:
$$\frac{\sqrt{(a-6)^2+b^2-36}}{\sqrt{(a+4)^2+\left(b-\frac{3}{2}\right)^2-\frac{73}{4}}}=\frac{2}{3}\ldots$$