Q. The circles $x^{2} + y^{2} - 4x - 6y - 12 = 0$ and $x^{2} + y^{2} + 4x + 6y + 4 = 0$
A) Touch externally
B) Do not intersect
C) Intersect at two points
D) Are concentric
Explanation
(c)
The centres and radii of given circles are $C_{1}(2,\ 3),\ C_{2}( - 2,\ - 3)$
and $r_{1} = \sqrt{4 + 9 + 12} = 5,\ \ \ r_{2} = \sqrt{4 + 9 - 4} = 3$
Now, $C_{1}C_{2} = \sqrt{(2 + 2)^{2} + (3 + 3)^{2}} = \sqrt{52}$
Here, $C_{1}C_{2} < r_{1} + r_{2}$
Hence, given circles intersect at two points
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.