MCQ
If the straight line $4x + 3y + \lambda = 0$ touches the circle $2({x^2} + {y^2}) = 5$, then $\lambda $ is
  • A
    $\frac{{5\sqrt 5 }}{2}$
  • B
    $5\sqrt 2 $
  • C
    $\frac{{5\sqrt 5 }}{4}$
  • $\frac{{5\sqrt {10} }}{2}$

Answer

Correct option: D.
$\frac{{5\sqrt {10} }}{2}$
d
(d) According to equation,

$\frac{{4(0) + 3(0) + \lambda }}{{\sqrt {{4^2} + {3^2}} }}$

$= \sqrt {\frac{5}{2}}$

$\Rightarrow \lambda = \frac{{5\sqrt 5 }}{{\sqrt 2 }} $

$= \frac{{5\sqrt {10} }}{2}$.

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