Q. Two circles $x^{2} + y^{2} - 2x - 3 = 0$ and $x^{2} + y^{2} - 4x - 6y - 8 = 0$ are such that
A) They touch internally
B) They touch externally
C) They intersect at two points
D) They are non-intersecting
Explanation
(c)
The centres of given circles are $C_{1}(1,\ 0),\ C_{2}(2,\ 3)$ and
$r_{1} = \sqrt{1^{2} + 0 + 3} = 2,\ \ r_{2} = \sqrt{4 + 9 + 8} = \sqrt{21}$
Now, $C_{1}C_{2} = \sqrt{(2 - 1)^{2} + (3 - 0)^{2}} = \sqrt{10} = 3.16$
and $r_{1} + r_{2} = 2 + \sqrt{21} = 6.58$
Hence, two circles intersect, each other at two points
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.