Question

Given the three circles $x^2+y^2-16x+60=0,3x^2+3y^2-36x+81=0$ and $x^2+y^2-16x-12y+84=0$. find (1) the point from which the tangents to them are equal in length and (2) this length.

Answer 1

Let the point from which the tangents are drawn to circles be $P(h,k)$

Given, length of tangents are equal

$(\sqrt{S_{11}}{)}_{C1}=(\sqrt{S_{11}}{)}_{C2}=(\sqrt{S_{11}}{)}_{C3}$

Equating $(\sqrt{S_{11}}{)}_{C1}=(\sqrt{S_{11}}{)}_{C2}$

$\sqrt{h^2+k^2-16h+60}=\sqrt{h^2+k^2-12h+27}60-27=4hh=\frac{33}{4}$

Equating $(\sqrt{S_{11}}{)}_{C2}=(\sqrt{S_{11}}{)}_{C3}$

$\sqrt{h^2+k^2-16h+60}=\sqrt{h^2+k^2-16h-12k+84}12k=84-60k=2$

Point $P(\frac{33}{4},2)$

Length $(\sqrt{S_{11}}{)}_{C1}$$=\sqrt{{\left(\frac{33}{4}\right)}^2+4-16\times \frac{33}{4}+60}=\sqrt{\frac{1}{16}}=\frac{1}{4}units$