Question

The equations to the tangents to the circle $x^2+y^2-6x+4y=12$ which are parallel to the straight line $4x+3y+5=0$, are?

Answer 1

For the circle ...$x²+y²-6x+4y-12=0,$

center $C\equiv (-6/-2,4/-2)\equiv (3,-2)$,

radius $r=\surd (9+4+12)=5$.

A tangent parallel to line $4x+3y+5=0$ has equation

$4x+3y+c=0$ .......... (1)

If seg CP is perpendicular to this tgt, then

$CP=r$

$r=\frac{|4\left(3\right)+3(-2)+c|}{\surd (4²+3²)}=5$

$|c+6|=25$

$c+6=\pm 25$

$=-6\pm 25$

$c=19,-31$........... (2)

From (1) and (2), the req'd tgt lines are

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