Question

Find the equation to the pair of transverse common tangents to the circles $x^2+y^2-4x-10y+28=0and{x}^2+y^2+4x-6y+4=0$

Answer 1

Equations of given circles are,

$S\equiv x^2+y^2-4x-10y+28=0$ ------ ( 1 )

$S\equiv x^2+y^2+4x-6y+4=0$ ------- ( 2 )

$Centers(-g,-f):$ $C_1(2,5),C_2(-2,3)$

$Radius\left(r\right)=\sqrt{g^2+f^2-c}$

$r_1=\sqrt{2^2+5^2-28}=\sqrt{4+25-28}=1$

$r_2=\sqrt{2^2+3^2-4}=\sqrt{4+9-4}=\sqrt{9}=3$

Internal center of similutute divides $C_1{C}_2$

Internally in the ratio $r_1:r_2=1:3=m:n$

$P=[\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}]=(\frac{1(-2)+3\left(2\right)}{1+3},\frac{1\left(3\right)+3\left(5\right)}{1+3})$

$=(\frac{4}{4},\frac{18}{4})=(1,\frac{9}{2})$

let $m$ be the slope of common tangent

Equation of tangent is $(y-y_1)m=(x-x_1)p(1,\frac{9}{2})$

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

$\Rightarrow$ $2mx-2y-2m+9=0$ ------- ( 1 )

Condition for tangent $r=\perp r$ distance

$r_1=\perp r$ distance from $C_1(2,5)$ to $\left(1\right)$

$d=\frac{|a{x}_1+b{y}_1+c|}{\sqrt{a^2+b^2}}\Rightarrow \frac{|2m\left(2\right)-2\left(5\right)-2m+9|}{\sqrt{(2m{)}^2+(2{)}^2}}=1$

$\Rightarrow$ $\frac{|2m-1|}{\sqrt{4m^2+4}}=1$ $S.O.B.S$

$\Rightarrow$ $(2m-1{)}^2=4m^2+4$

$\Rightarrow$ $4m^2+1-4m=4m^2+4$ $\{m=\infty =\frac{1}{0}$

$\Rightarrow$ $-3=4m$ or $m=\frac{-3}{4}$

$\therefore$ Equation of tangents are,

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

And $(y-\frac{9}{2})=-\frac{3}{4}(x-1)$

$\Rightarrow$ $4y-18=-3x+3$

$\Rightarrow$ $3x+4y-21=0$

$3x+4y-21=0$ and $x-1=0$