Answer:
Option C
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
