Q. Two diameters of the circle $3x^{2} + 3y^{2} - 6x - 18y - 7 = 0$ are along the lines $3x + y = c_{1}$and$x - 3y = c_{2}$. Then, the value of $c_{1}c_{2}$ is
A) $- 48$
B) 80
C) $- 72$
D) 54
Explanation
(a)
Given, $x^{2} + y^{2} - 2x - 6y - \frac{7}{3} = 0$
The centre of this circle is (1, 3)
Also, two diameter of this circle are along the lines $3x + y = c_{1}$ and $x - 3y = c_{2}$
These two diameters should be passed from (1, 3)
$\therefore\ \ c_{1} = 6$ and $c_{2} = - 8$
Hence, $c_{1}c_{2} = 6 \times ( - 8) = - 48$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.