Question

For circles $(x-3{)}^2+(y-2{)}^2=9$ and $(x+5{)}^2+(y+6{)}^2=9$ which among the following are tangent equations

• A. $y+2=\frac{16-3\sqrt{23}}{7}(x+1)$
• B. $y+2=\frac{16+3\sqrt{23}}{7}(x+1)$
• C. $y+1=x+3\sqrt{2}$
• D. $y+1=x-3\sqrt{2}$
  

Answer 1

Answer: (A), (B), (C), (D)

$y+1=x-3\sqrt{2}$


$C_1\to (3,2)r_1=3$

$C_2\to (-5,-6)r_2=3$


Internal point of simlitude $\left(I\right)$ divide $C_1$ and $C_2$ in $1:1$ ratio internally
​​​​​​so, $I\equiv (-1,-2)$
Let the equation of tangents passing through $I$ be

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

Perpendicular distance from $C_1$ to the tangent = $3$
$\Rightarrow \left|\frac{4m-4}{\sqrt{1+m^2}}\right|=3\Rightarrow 16(m-1{)}^2=9(m^2+1)\Rightarrow 7m^2-32m+7=0\Rightarrow m=\frac{16\pm 3\sqrt{23}}{7}$
So, equation of Transverse common tangents are
$y+2=\frac{16\pm 3\sqrt{23}}{7}(x+1)$
Radius of both circles are same.

Direct Common tangent are parallel to $C_1{C}_2$, so, D.C.T. equation will be of the form
$y=x+\lambda (\because m_{C_1{C}_2}=1)$
Perpendicular distance = radius
$\Rightarrow \left|\frac{1+\lambda }{\sqrt{2}}\right|=3$

$\Rightarrow \lambda +1=\pm 3\sqrt{2}$
$\Rightarrow \lambda =\pm 3\sqrt{2}-1$

So, equation of D.C.T. are $y=x+3\sqrt{2}-1$ and $y=x-3\sqrt{2}-1$