Question

For the circles $S_1\equiv x^2+y^2-4x-6y-12=0$ and $S_2\equiv x^2+y^2+6x+4y-12=0$ and the line $L\equiv x+y=0$

• A. $L$ is the common tangent of $S_1$ and $S_2$
• B. $L$ is the common chord of $S_1$ and $S_2$
• C. $L$ is radical axis of $S_1$ and $S_2$
• D. $L$ is perpendicular to the line joining the centers of $S_1$ and $S_2$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $L$ is radical axis of $S_1$ and $S_2$
C $L$ is perpendicular to the line joining the centers of $S_1$ and $S_2$
D $L$ is the common chord of $S_1$ and $S_2$

Centre of $S_1$ is $C_1\equiv (2,3)$ and radius of $S_1$ is $r_1=5.$

Centre of $S_1$ is $C_2\equiv (-3,-2)$and radius of $S_1$ is $r_1=5.$

Also, $d=$ distance between centres $=C_1{C}_2=\sqrt{25+25}=\sqrt{50}$

$\therefore |r_1-r_2|=0$ and $r_1+r_2=10.$

Since $|r_1-r_2|<C_1{C}_2<r_1+r_2,$ therefore circles $S_1$ and $S_2$ intersect.

Equation of the common chord of $C_1$ and $C_{21}$ is $S_1-S_2=0$ i.e., $x+y=0,$

which is also the equation of the radical axis and hence it is $\perp$ to the line joining the centres $C_1$ and $C_2$ of the two circles.