Question

If the lengths of the tangents from a point to the circles
$S_1:x^2+y^2-16x+60=0,$
$S_2:x^2+y^2-12x+20=0,$
$S_3:x^2+y^2-16x-12y=0$ , are equal and its value is $L$, then find $L$.

Answer 1

$S_1:x^2+y^2-16x+60=0$
$S_2:x^2+y^2-12x+20=0$
$S_3:x^2+y^2-16x-12y=0$
Locus of point from which we can draw tangents of equal length to three circle is radical centre that is a point of intersection of radical axes.
$\therefore S_1-S_2=0$ obtains radical axis
$\Rightarrow -4x+40=0$
$x=10...\left(1\right)$
$S_2-S_3=0$ obtains another radical axis
$4x+12y+20=0$
$x+3y+5=0...\left(2\right)$
From (1) & (2), we get
$x=10,y=-5$
Thus, radical centre is $(10,-5)$
Therefore, length of tangent from $(10,-5)$ to $S_1$ is
$L=\sqrt{S_1}=\sqrt{{10}^2+5^2-160+60}$
$L=\sqrt{25}=5$