The centre(s) of the circle(s) passing throug | vedclass

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Question

consider family of $ \odot$ 's through $(0, 0)$ and $(1, 0)$

$x(x - 1) + y^2 + \lambda y = 0$

touches $x^2 + y^2 = 9$

$	herefore$ common

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$(B)$ or $(C)$ both

Vedclass · Answer

consider family of $ \odot$ 's through $(0, 0)$ and $(1, 0)$

$x(x - 1) + y^2 + \lambda y = 0$

touches $x^2 + y^2 = 9$

$	herefore$ common chord $= - x + hy + 9 = 0....(1)$

$	herefore$ perpendicular from $(0, 0)$ on $(1)$ is equal to $3$.

$\left| {\,\frac{9}{{\sqrt {1 + {\lambda ^2}} }}\,} ight|$ $= 3$ $\Rightarrow \,\, \lambda_2 = 8$ $\Rightarrow \,\, \lambda = \pm 2\sqrt{2}$

circle $x (x - 1) + y^2 + 2\sqrt{2} y =0$

$	herefore$    centre $\left( {\frac{1}{2}\,\,,\,\, - \,{2^{1/2}}} ight)$

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 :

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