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The centre$(s)$ of the circle$(s)$ passing through the points $(0, 0) , (1, 0)$ and touching the circle $x^2 + y^2 = 9$ is/are :
$\left( {\frac{3}{2}\,\,,\,\,\frac{1}{2}} \right)$
$\left( {\frac{1}{2}\,\,,\,\, - \,{2^{1/2}}} \right)$
$\left( {\frac{1}{2}\,\,,\,\,{2^{1/2}}} \right)$
$(B)$ or $(C)$ both
Solution
consider family of $ \odot$ 's through $(0, 0)$ and $(1, 0)$
$x(x – 1) + y^2 + \lambda y = 0$
touches $x^2 + y^2 = 9$
$\therefore$ common chord $= – x + hy + 9 = 0….(1)$
$\therefore$ perpendicular from $(0, 0)$ on $(1)$ is equal to $3$.
$\left| {\,\frac{9}{{\sqrt {1 + {\lambda ^2}} }}\,} \right|$ $= 3$ $\Rightarrow \,\, \lambda_2 = 8$ $\Rightarrow \,\, \lambda = \pm 2\sqrt{2}$
circle $x (x – 1) + y^2 + 2\sqrt{2} y =0$
$\therefore$ centre $\left( {\frac{1}{2}\,\,,\,\, – \,{2^{1/2}}} \right)$
