Question

For circles $x^2+y^2+2x-8y+13=0$ and $x^2+y^2-12x-14y+76=0$ equation of all the common tangents are:

• A. $y+2=\frac{21\pm \sqrt{57}}{48}(x+15)$
• B. $y-\frac{26}{5}=\frac{21\pm 5\sqrt{33}}{24}(x-\frac{9}{5})$
• C. $y-\frac{26}{5}=\frac{21\pm 5\sqrt{33}}{24}(x-\frac{10}{5})$
• D. $y-2=\frac{21\pm \sqrt{57}}{48}(x-15)$

Answer 1

Answer: (A), (B)

$y-\frac{26}{5}=\frac{21\pm 5\sqrt{33}}{24}(x-\frac{9}{5})$

Circle $1$: Centre $C_1(-1,4),$ radius=$2$
Circle $2$: Centre $C_2(6,7),$ radius=$3$
Consider, $|C_1{C}_2|=\sqrt{58}$ and $r_1+r_2=5\Rightarrow |C_1{C}_2|>r_1+r_2$
$\therefore 4$ common tangents exists:$2$TCTs and $2$DCTs
$Q$ divides the join of $C_1,C_2$ internally in the ratio $r_1:r_2$
$\therefore C_{1}Q:C_{2}Q=2:3$
$\Rightarrow Q=(\frac{9}{5},\frac{26}{5})$
Thus by point slope form
Equation of transverse common tangents passing through $Q=(\frac{9}{5},\frac{26}{5})$ is:
$\Rightarrow 5mx-5y+26-9m=0$
But this touches both the circle
$\Rightarrow$ pependicular distance=radius
$\Rightarrow \left|\frac{-5m+6-9m}{\sqrt{25m^2+25}}\right|=2$
Solving this we get,
$m=\frac{21\pm 5\sqrt{33}}{24}$
Hence equation of transverse common tangent $T_3,T_4$ are:
$y-\frac{26}{5}=\frac{21\pm 5\sqrt{33}}{24}(x-\frac{9}{5})$
$p$ divides the join of $C_1,C_2$ externally in the ratio $r_1:r_2$
$\therefore P{C}_1:P{C}_2=2:3$
$\Rightarrow P=(-15,-2)$
Equation of direct common tangents passing through $P(-15,-2)$ is:
$y+2=m(x+15)$
$\Rightarrow mx-y+15m-2=0$
$\Rightarrow$ perpendicular distance = radius
$\Rightarrow \left|\frac{-m-4+15m-2}{\sqrt{m^2+1}}\right|=2$
Solving this,we get,
$m=\frac{21\pm \sqrt{57}}{48}$
Hence equation of direct common tangents $T_1,T_2$ are:
$y+2=\frac{21\pm \sqrt{57}}{48}(x+15)$