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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Statement $I$ : For any two non-zero complex numbers $\mathrm{z}_1, \mathrm{z}_2$
$\left(\left|z_1\right|+\left|z_2\right|\right)\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq 2\left(\left|z_1\right|+\left|z_2\right|\right)$ and
Statement $II$ : If $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are three distinct complex numbers and a, b, c are three positive real numbers such that $\frac{a}{|y-z|}=\frac{b}{|z-x|}=\frac{c}{|x-y|}$, then
$\frac{\mathrm{a}^2}{\mathrm{y}-\mathrm{z}}+\frac{\mathrm{b}^2}{\mathrm{z}-\mathrm{x}}+\frac{\mathrm{c}^2}{\mathrm{x}-\mathrm{y}}=1$
Between the above two statements,