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. $3$
• B. $4$
• C. $1$
• D. $2$

Answer 1

The correct option is A $3$

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

Compare with $x^2+y^2+2gx+2fy+c=0$

centre $=(-g,-f)$$=(2,3)$

Radius $=\sqrt{g^2+f^2+c}=\sqrt{2^2+3^2+12}$

$=\sqrt{4+9+12}$

$=\sqrt{25}=5$

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

Centre $=(-g,-f)$ $=(-3,-9)$

Radius $=\sqrt{g^2+f^2-c}$

$=\sqrt{3^2+9^2-26}$ $=\sqrt{9+81-26}$

$=\sqrt{64}=8$

The Distance between centre

$=\sqrt{(2+3{)}^2+(3+9{)}^2}$

$=\sqrt{5^2+{12}^2}$

$=13$

Sum of radius $=8+5=13$.

As we can see that Distance between the centres=sum of the radius.

Hence, the circles are touching each other so the number of tangents $=3$