Q. The values of $\lambda$ so that the line $3x - 4y = \lambda$ touches $x^{2} + y^{2} - 4x - 8y - 5 = 0$ are
A) $- 35,\ 15$
B) $3,\ - 5$
C) $35,\ - 15$
D) $- 3,\ 5$
Explanation
(a)
The centre of circle is (2, 4)
Radius$= \sqrt{4 + 16 + 5} = 5$
$\because$ Perpendicular distance of $3x - 4y - \lambda = 0$ from (2, 4) is equal to the readius of circle
$\therefore\ \ \left| \frac{6 - 16 - \lambda}{\sqrt{9 + 16}} \right| = 5$
$\Rightarrow \ - 10 - \lambda = \pm 25\ \ \ \ \ \lambda = - 35,\ 15$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.