Let the centre of a circle, passing through t | vedclass

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Question

$Image$

$ (\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2=\mathrm{h}^2+\mathrm{k}^2 $

$ \mathrm{x}^2+\mathrm{y}^2-2 \mathrm{hx}-2 \mathrm{ky

Vedclass · Accepted Answer

$9$

Vedclass · Answer

$Image$

$ (\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2=\mathrm{h}^2+\mathrm{k}^2 $

$ \mathrm{x}^2+\mathrm{y}^2-2 \mathrm{hx}-2 \mathrm{ky}=0 $

$ \because 	ext { passes through }(1,0) $

$ \Rightarrow 1+0-2 \mathrm{~h}=0 $

$ \Rightarrow \mathrm{h}=1 / 2 $

$ \because \mathrm{OC}=\frac{\mathrm{OP}}{2} $

$ \sqrt{\left(\frac{1}{2}ight)^2+\mathrm{k}^2}=\frac{3}{2}$

$ \frac{1}{4}+\mathrm{k}^2=\frac{9}{4} $

$ \mathrm{k}^2=2 $

$ \mathrm{k}= \pm \sqrt{2}$

$	herefore$ Possible coordinate of

$ \mathrm{c}(\mathrm{h}, \mathrm{k})\left(\frac{1}{2}, \sqrt{2}ight)\left(\frac{1}{2},-\sqrt{2}ight) $

$ 4\left(\mathrm{~h}^2+\mathrm{k}^2ight)=4\left(\frac{1}{4}+2ight)=4\left(\frac{9}{4}ight)=9$

Let the centre of a circle, passing through the point $(0,0),(1,0)$ and touching the circle $x^2+y^2=9$, be $(h, k)$. Then for all possible values of the coordinates of the centre $(h, k), 4\left(h^2+k^2\right)$ is equal to .............

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