Question

Find the equations of tangents to the circle $x^2+y^2-6x+4y-12=0$ which are parallel to the line $4x-3y+6=0$

• A. $4x+3y+7=0$ and $4x+3y-43=0$
• B. $4x-3y+7=0$ and $4x-3y-43=0$
• C. $3x-4y+7=0$ and $3x-4y+43=0$
• D. $3x-4y+7=0$ and $3x-4y-43=0$

Answer 1

The correct option is B $4x-3y+7=0$ and $4x-3y-43=0$

Centre and radius of the given circle is $(3,-2)$ and $\sqrt{9+4+12}=5$ respectively.

Equation of line parallel to $4x-3y+6=0$ is given by,

$4x-3y+k=0$. Given this line is tangent to the given circle.

perpendicular distance from center to tangent is radius

$\Rightarrow \left|\frac{4\left(3\right)-3(-2)+k}{\sqrt{3^2+4^2}}\right|=5$

$\Rightarrow 18+k=\pm 25\Rightarrow k=7,-43$

Hence required line are, $4x-3y+7=0$ and $4x-3y-43=0$