Question

Equation of tangent to the circle $x^2+y^2-6x+4y-12=0$ which are parallel to the line $4x+3y+5=0$ is ?

• A. $4x+3y+19=0$
• B. $4x+3y+31=0$
• C. $4x+3y-19=0$
• D. $4x+3y+29=0$

Answer 1

The correct option is A $4x+3y+19=0$

For the circle $x²+y²-6x+4y-12=0$,

center $C\equiv (\frac{-6}{-2},\frac{4}{-2})=(3,-2)$

radius $r=\sqrt{(9+4+12)}=5$

A tangent parallel to line $4x+3y+5=0$ has equation $4x+3y+c=0⟹\left(1\right)$

If seg $CP$ is perpendicular to this tgt,then

$CP=r$

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

$|c+6|=25$

$c+6=\pm 25$

$c=-6\pm 25$

$c=19,-31⟹\left(2\right)$

From (1) and (2)

$4x+3y+19=0and4x+3y-31=0.$