Question

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:

• A. $2$
• B. $3$
• C. $4$
• D. $1$

Answer 1

The correct option is A $2$

$x^2+y^2-4x-6y-12=0$

${(x-2)}^2+{(y-3)}^2-4-9-12=0$

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

Here,

Centre of circle $\left(C_1\right)\equiv (2,3)$

Radius of circle $\left(r_1\right)\equiv 5$

$x^2+y^2+6x+18y+26=0$

${(x+3)}^2+{(y+9)}^2-9-81+26=0$

${(x-2)}^2+{(y-3)}^2={\left(8\right)}^2$

Here,

Centre of circle $\left(C_2\right)\equiv (-3,-9)$

Radius of circle $\left(r_2\right)\equiv 8$

Now,

$C_1{C}_2=\sqrt{{(-3-2)}^2+{(-9-3)}^2}$

$\Rightarrow C_1{C}_2=\sqrt{25+144}=13$

$r_1+r_2=5+8=13$

$\because C_1{C}_2=r_1+r_2$

Therefore,

No. of common tangents $=3$

Hence the correct answer is $3$.