Discussion Forum

 The number of common tangents to  the circles   $x^{2}+y^{2}-4x-6y-12=0$  and $x^{2}+y^{2}+6x+18y+26 =0$  is

Answer:

Explanation:

Central Idea:

  Number of common tangents depend on the position of the circle with respect to each other.

  (i) If circles touch externally   $\Rightarrow$    C₁ C₂  =r₁+r₂   , 3 common tangents

  (ii)  If circles touch internally  $\Rightarrow$   C₁ C₂  =r₂ -r₁ , 1 common integer

   (iii) If circles do not touch each other, 4 common tangents

 Given equations of circles are

      $x^{2}+y^{2}-4x-6y-12=0$      ......(i)

      $x^{2}+y^{2}+6x+18y+26 =0$     .......(ii)

  Centre of circle (i) C₁   (2,3) and radius 

                           = $\sqrt{4+9+12}=5(r_{1})$    (say)

 Centre of circle (ii) is C₂  (-3,-9)  and radius

                     =$\sqrt{9+81-26}=8(r_{2})$               (say)

Now, C₁ C₂=   $\sqrt{(2+3)^{2}+(3+9)^{2}}$

$\Rightarrow$     $C_{1}C_{2}=\sqrt{5^{2}+12^{2}}$

  $\Rightarrow$    $C_{1}C_{2}=\sqrt{25+144}=13$

 $\therefore$      $r_{1}+r_{2}=5+8=13$

  Also, C₁ C₂  = r₁ +r₂

  Thus, both circles touch each other externally. Hence there are three common tangents