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$\frac{{5\sqrt 5 }}{4}$
- ✓$\frac{{5\sqrt {10} }}{2}$
$\frac{{4(0) + 3(0) + \lambda }}{{\sqrt {{4^2} + {3^2}} }}$
$= \sqrt {\frac{5}{2}}$
$\Rightarrow \lambda = \frac{{5\sqrt 5 }}{{\sqrt 2 }} $
$= \frac{{5\sqrt {10} }}{2}$.
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$(A)$ $a+b=4$
$(B)$ $a-b=2$
$(C)$ The length of the diagonal $P R$ of the parallelogram $P Q R S$ is $4$
$(D)$ $\overrightarrow{ w }$ is an angle bisector of the vectors $\overrightarrow{ PQ }$ and $\overrightarrow{ PS }$