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Q.
Consider the circle $x^2 + y^2 - 6\,x + 4\,y = 12$. The equation of a tangent to this circle that is parallel to the line $4\,x + 3\,y + 5 = 0$ is
TS EAMCET 2017
Solution:
We have,
$ x^{2}+y^{2}-6 \,x+4\, y=12 $
$\Rightarrow (x-3)^{2}+(y+2)^{2}=25 $
$\Rightarrow x-3)^{2}+(y+2)^{2}=(5)^{2} $
Equation of tangent whose slope $m$ is
$y+2=m(x-3) \pm 5 \sqrt{m^{2}+1}\,.....(i)$
Now, this tangent is parallel to line $4 \,x+3\, y+5=0$
$\therefore $ Slope of line is $-\frac{4}{3}$
Put the value of $m=-\frac{4}{3}$ is Eq. (i), we get
$y+2=-\frac{4}{3}(x-3) \pm 5 \sqrt{\left(\frac{-4}{3}\right)^{2}+1}$
$\Rightarrow y+2=\frac{-4}{3}(x-3) \pm 5\left(\frac{5}{3}\right)$
$\Rightarrow 3 \,y+6=-4 \,x+12 \,\pm \,25$
$\Rightarrow 4 \,x+3 \,y=6 \,\pm \,25$
$\Rightarrow 4 \,x+3\, y=31$ or $4\,x+3 \,y=-19$
Hence, equation of tangent is
$4\, x+3 \,y-31=0$ or $4 \,x+3 \,y+19=0$