Q. The diameter of a circle are along $2x + y - 7 = 0$and$x + 3y - 11 = 0$. Then, the equation of this circle, which also passes through (5, 7), is
A) $x^{2} + y^{2} - 4x - 6y - 16 = 0$
B) $x^{2} + y^{2} - 4x - 6y - 20 = 0$
C) $x^{2} + y^{2} - 4x - 6y - 12 = 0$
D) $x^{2} + y^{2} + 4x + 6y - 12 = 0$
Explanation
(c)
The intersection point of diameter lines is (2, 3) which is the centre of circle
Now, radius$= \sqrt{(5 - 2)^{2} + (7 - 3)^{2}}$
$= \sqrt{9 + 16} = 5$
$\therefore$Required equation of circle is
$(x - 2)^{2} + (y - 3)^{2} = 5^{2}$
$\Rightarrow \ \ \ x^{2} + y^{2} - 4x - 6y - 12 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.