Question

The point where the line $4x-3y+7=0$ touches the circle $x^2+y^2-6x+4y-12=0$ is

• A. $(1,1)$
• B. $(1,-1)$
• C. $(-1,1)$
• D. $(-1,-1)$

Answer 1

Answer: (C)

$(-1,1)$

It is given that $4x-3y+7=0$ is a tangent to the circle $x^2+y^2-6x+4y-12=0$. The centre of the circle $C$ is $(3,-2)$. The line will touch the circle at the foot of the perpendicular from the centre (property of tangent).

Foot of the perpendicular from $C$ to tangent
$\frac{h-3}{4}=\frac{k+2}{-3}=-\frac{4\left(3\right)-3(-2)+7}{\sqrt{4^2+3^2}}$

$\Rightarrow \frac{h-3}{4}=-\frac{25}{25}$ and $-\frac{k+2}{3}=-\frac{25}{25}$

$\Rightarrow h-3=-4$ and $k+2=3$
$\Rightarrow h=-1$ and $k=1$

Hence the point of contact is $(-1,1)$