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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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