Q85 / 160 · WBJEE 2009 · Mathematics

The circles $\mathrm{x}^2+\mathrm{y}^2-10 \mathrm{x}+16=0$ and $\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2$ intersect at two distinct points if

  • A
    $\mathrm{a} < 2$
  • Correct answer
    B
    $2 < $ a $ < 8$
  • C
    $a>8$
  • D
    $\mathbf{a}=2$

Solution

Hints : $C_1(5,0) \quad r_1=\sqrt{25-16}=3$ $$ \begin{aligned} & \mathrm{C}_2(0,0) \quad \mathrm{r}_2=\mathbf{a} \\ & \mathrm{r}_1 \& \mathrm{r}_2 < \mathrm{C}_1 \mathrm{C}_2 < \mathrm{r}_1+\mathrm{r}_2 \\ & |\mathbf{a}-3| < \sqrt{25} < \mathrm{a}+3 \\ & |\mathbf{a}-3| < 5 < \mathrm{a}+3 \\ & -5 < \mathrm{a}-3 < 5 \\ & -2 < \mathrm{a} < 8 \\ & 2 < \mathrm{a} < 8 \end{aligned} $$

Hence, the correct answer is option: 'B'.

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