Question

The tangent lines to the circle $x^2+y^2-6x+4y=12$ which are parallel to the line $4x+3y+5=0$ are given by

• A. $4x+3y+5=0,4x+3y-25=0$
• B. $4x+3y-31=0,4x+3y+19=0$
• C. $4x+3y-17=0,4x+3y+13=0$
• D. None of these

Answer 1

The correct option is A $4x+3y+5=0,4x+3y-25=0$

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

Centre of circle $=(-g,-f)=(3,-2)$

and radius $a=\sqrt{g^2+f^2-c}=\sqrt{9+14+12}$

$=5$

So $e{q}^n$ of tangent $y=mx\pm \sqrt{1+m^2}$

So $y=mx\pm 5\sqrt{1+m^2}....\left(1\right)$

but here tangent is parallel to the line $4x+3y+5=0$

$3y=-5-4x$

$y=\frac{-5}{3}-\frac{4}{3}x$

So here slope $=\frac{-4}{3}$

So from (i) tangent $e{q}^n$

$y=\frac{-4x}{3}\pm 4\sqrt{1+\frac{16}{9}}$

$y=\frac{-4x}{3}\pm 5\times \frac{5}{3}$

$3y=-4x\pm 25\Rightarrow$ so here $3y+4x+25=0$

or $3y+4x-25=0$