Question

If two circles of equal radii of $5$ unit each, touch externally at $(1,2)$ and the equation of one common tangent is $3x-4y+30=0,$ then the equations of the other two common tangents are

• A. $3x-4y-20=0,4x+3y-5=0$
• B. $3x-4y+20=0,4x+3y+10=0$
• C. $3x-4y-20=0,4x+3y-10=0$
• D. $3x-4y+20=0,4x+3y-10=0$

Answer 1

The correct option is C $3x-4y-20=0,4x+3y-10=0$


Let $L_1,L_2$ be the other two tangents.
$L_1$ is $\perp$ to $3x-4y+30=0$ and passing through $(1,2)$
$\therefore 4x+3y+k=0$
$\Rightarrow 4+6+k=0$
$\Rightarrow k=-10$
Hence, equation of $L_1$ is $4x+3y-10=0$

Since, both the circles have equal radius,
therefore, $L_2$ is parallel to $3x-4y+30=0$ and distance between them is $10$
Let equation of $L_2$ be $3x-4y+l=0$
So, $\frac{|30-l|}{\sqrt{3^2+4^2}}=10$
$\Rightarrow 30-l=\pm 50$
$\Rightarrow l=-20,80$
Equation of $L_2$ is $3x-4y-20=0$ or $3x-4y+80=0$
As distance between $L_2$ and $(1,2)$ is $5$ units,
$\frac{|3-8-20|}{\sqrt{25}}=5$
and $\frac{|3-8+80|}{\sqrt{25}}=15$
Hence, $L_2$ is $3x-4y-20=0,$ and $L_1$ is $4x+3y-10=0$