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A circle passing through the points $(3, \sqrt{7 / 2})$ touches the pair of lines $\mathrm{x^{2}-y^{2}-2 x+1=0}$ . The centre of the circle is
Option: 1
$(4,0)$

Option: 2
$\quad(7,0)$

Option: 3
$(0,4)$

Option: 4
$(5,0)$

Answers (1)

best_answer

Given pair of lines are $\mathrm{(x-1)^{2}=y^{2}}$
$\mathrm{\Rightarrow \mathrm{x}-1=\mathrm{y}\: and \: \mathrm{x}+\mathrm{y}-1=0}$

The centre will lie on x-axis $\mathrm{\left ( \alpha ,0 \right )}$

$\left|\frac{\alpha-1}{\sqrt{2}}\right| \sqrt{(\alpha-3)^{2}+\frac{7}{2}}$
$\frac{\alpha^{2}+1-2 \alpha}{2}=(\alpha-3)^{2}+\frac{7}{2}$
$\frac{\alpha^{2}+1-2 \alpha}{2}=(\alpha-3)^{2}+\frac{7}{2}$
$\alpha^{2}+1-2 \alpha=2 \alpha^{2}-12 \alpha+18+7$
$\alpha^{2}-10 \alpha+24=0$
$\alpha^{2}-6 \alpha-4 \alpha+24=0$
$(\alpha-6)(\alpha-4)=0$
$\alpha=4,6$
Hence (A) is the correct answer.

Posted by

Divya Prakash Singh

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