Question

The lengths of the tangents from $P(1,-1)$ and $Q(3,3)$ to a circle are $\sqrt{2}$ and $\sqrt{6}$ , respectively. Then, find the length of the tangent from $R(-1,-5)$ to the same circle.

Answer 1

Length of tangent is $\sqrt{x^2+y^2+2yx+2hy+f}$ drawn from $(x,y)$

here $(x,y)=(1,-1)$ & $(3,3)$

$\sqrt{(1{)}^2+(-1{)}^2+2\left(1\right)g+2(-1)h+F}=\sqrt{2}$

$1+1+2y-2h+f=2$

$2g-2h+f=0$ ………..$\left(1\right)$

$\sqrt{(3{)}^2+(3{)}^2+2\left(3\right)g+2\left(3\right)h+f}=\sqrt{6}$

$9+9+6g+6h+f=6$

$6g+6h+f=-12$

$2g+2h+\frac{f}{3}=-4$ ………….$\left(2\right)$ divide by $3$

Subtract $\left(1\right)$ from $\left(2\right)$

$\frac{f}{3}-f=-4$

$-\frac{2f}{3}=-4$

$f=12$ put in $\left(1\right)$ & $\left(2\right)$

$2g-2h=-12$

$\Rightarrow g-h=-6$ ………$\left(3\right)$

$2g+2h=-8$

$\Rightarrow g+h=-4$ ………..$\left(4\right)$

$Eq.\left(3\right)+\left(4\right)$

$2g=-10$

$\Rightarrow g=-5$

$h=9+6=-5+6=1$

Length of tangent from $(-1,-5)$

$\sqrt{(-1{)}^2+(-5{)}^2+2(-1)(-5)+2(-5)\left(1\right)+12}$

$=\sqrt{1+5+10-10+12}=\sqrt{18}$

Length$=\sqrt{18}=3\sqrt{2}$.