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

Third equation in a misprint. It must actually be :

$x^2+y^2-16x-12y+84=0$

Length of tangent $=\sqrt{s_1}$

Let the required point be $(h,k)$

Equation of circles:

x^{2}+y^{2}-16x+60=0$

$3(x^2+y^2-12x+27)=0$

$\Rightarrow x^2+y^2-12x+27=0$

and $x^2+y^2-16x-12y+84=0$

Given that length of tangent on all three circles from $(h,k)$ are equal ibn length.

$\Rightarrow \sqrt{h^2+k^2-16h+60}=\sqrt{h^2+k^2-12h+27}$

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

$\Rightarrow h^2+k^2-16h+60=h^2+k^2-12h+27$

$\Rightarrow 4h=33$

$\Rightarrow h=\frac{33}{4}$

and

$h^2+k^2-16h+60=h^2+k^2-16h-12k+84$

$\Rightarrow 12k=24$

$\Rightarrow k=2$

$\therefore (\frac{33}{4},2)$ is the point from which tangents drawn are equal in length.

Length of tangent $=\sqrt{(\frac{33}{4}{)}^2+(2{)}^2-16\left(\frac{33}{4}\right)-12\left(2\right)+84}$

$\sqrt{\frac{1089}{16}+4-132+60}$

$=\sqrt{\frac{1089}{16}+64-132}$

$=\sqrt{\frac{1089}{16}-68}$

$=\sqrt{\frac{1}{16}}=\frac{1}{4}$