Discussion Forum

Consider the circle $x^{2}+y^{2}-6x+4y=12$. The equations of a tangent of this circle that is parallel to the line x+3y+5=0

Answer:

Explanation:

We have ,

$x^{2}+y^{2}-6x+4y=12$.

$\Rightarrow$  $(x-3)^{2}+(y+2)^{2}=25$

$\Rightarrow$     $(x-3)^{2}+(y+2)^{2}=5^{2}$

 Equation of tangent  whose slope m is 

  $y+2=m(x-3)\pm5\sqrt{m^{2}+1}$   .......(i)

 Now, this tangent is parallel to line 4x+3y+5=0

 $\therefore$  slope of line is -$\frac{4}{3}$

 Put the value of m= -$\frac{4}{3}$ is Eq.(i) , we get

   $y+2=-\frac{4}{3} (x-3)\pm 5\sqrt{\left(\frac{-4}{3}\right)^{2}+1}$

$\Rightarrow$      $y+2=\frac{-4}{3} (x-3)\pm 5\left(\frac{5}{3}\right)$

$\Rightarrow$    $3y+6=-4x+12\pm25$

$\Rightarrow$   $4x+3y=6 \pm 25$

$\Rightarrow$   $4x+3y=31$ or $4x+3y=-19$

Hence , equation  of tangent is 

   $4x+3y-31=0$ or  $4x+3y+19=0$