Question

The equation of the tangents to the circle $x^2+y^2-4x-6y-12=0$, which are perpendicular to the line $4x+3y=7$ are

• A. $3x+4y+19=0,3x+4y+31=0$
• B. $4x-3y+19=0,4x-3y-31=0$
• C. $3x-4y+31=0,3x-4y-19=0$
• D. none of these

Answer 1

Answer: (C)

$3x-4y+31=0,3x-4y-19=0$

$x^2+y^2-4x-6y-12=0$

$C(-g,-f)=(2,3)$

$r=\sqrt{g^2+f^2-c}=\sqrt{2^2+3^2+12}=5$

Tangent perpendicular to the line $4x+3y=7$ has equation

$3x-4y+c=0$

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

$\left|\frac{6-12+c}{5}\right|=5$

$|-6+c|=25$

$-6+c=\pm 25$

$\therefore c=31,19$

$3x-4y+31=0,3x-4y+19=0$