Question

Find the transverse common tangents of the circle $x^2+y^2-4x-10y+28=0$ and $x^2+y^2+4x-6y+4=0$.

Answer 1

$x^2+y^2-4x-10y+28=0$ $x^2+y^2+4x-6y+4=0$

centre $c_1=(2,5)$ centre $c_2=(-2,3)$

$r_1=\sqrt{4+25-28}$ $r_2=\sqrt{4+9-4}$

$r_1=1$ $r_2=3$

Distance between $c_1$ and $c_2=\sqrt{(2+2{)}^2+{(5-3)}^2}=\sqrt{20}$

$\therefore c_1{c}_2>r_1{+r}_2$ ............ $[\sqrt{2}>4]$

$\therefore$ Point $P$ divides $c_1{c}_2$ in the ratio $1:3$

P = $[2+\frac{(-2-2)}{4},5+\frac{(3-5)}{4}]$

P = $(1,\frac{9}{2})$

Eqn of tangent with slope m is

$y-\frac{9}{2}=m(x-1)$ $\Rightarrow$ $2mx-2y+9-2m=0$

The above line is tangent to circle $x^2+y^2-4x-10y+28=0$

Radius $=$ bot distance from centre (2,5) to line

1 $=$ $\left|\frac{2m\left(2\right)-2\left(5\right)+9-2m}{\sqrt{4m^2+4}}\right|$ $\Rightarrow$ 1 = $\left|\frac{2m-1}{\sqrt{4m^2+4}}\right|$

$4m^2+4=4m^2+1-4m$

$4m+3=0$ $\Rightarrow$ $m=-\frac{3}{4}$

Since we have two tangents passing from point $(1,\frac{9}{2})$

But $m^2$ term is eliminated

$\therefore$ slope of other line is $\infty$

Eqn. of tangent having slope $-\frac{3}{4}$ is

$y-\frac{9}{2}=-\frac{3}{4}(x-1)$ $\Rightarrow$ $3x+4y-21=0$

eqn. of tangent having slope $\infty$ & passing though $(1,\frac{9}{2})$ is

$x=1$