Let the centre of a circle, passing through the points $(0,0),(1,0)$ 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_________

Question

Aicharya · Accepted Answer

$\begin{aligned} & (\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2=\mathrm{h}^2+\mathrm{k}^2 \\ & \mathrm{x}^2+\mathrm{y}^2-2 \mathrm{hx}-2 \mathrm{ky}=0 \\ & \because \text { passes through }(1,0) \\ & \Rightarrow 1+0-2 \mathrm{~h}=0 \\ & \Rightarrow \mathrm{h}=1 / 2 \\ & \because \mathrm{OC}=\frac{\mathrm{OP}}{2} \\ & \sqrt{\left(\frac{1}{2}\right)^2+\mathrm{k}^2}=\frac{3}{2}\end{aligned}$ $\begin{aligned} & \frac{1}{4}+\mathrm{k}^2=\frac{9}{4} \\ & \mathrm{k}^2=2 \\ & \mathrm{k}= \pm \sqrt{2} \end{aligned}$ $\therefore$ Possible coordinate of $\begin{aligned} & \mathrm{c}(\mathrm{h}, \mathrm{k})\left(\frac{1}{2}, \sqrt{2}\right)\left(\frac{1}{2},-\sqrt{2}\right) \\ & 4\left(\mathrm{~h}^2+\mathrm{k}^2\right)=4\left(\frac{1}{4}+2\right)=4\left(\frac{9}{4}\right)=9 \end{aligned}$

Let the centre of a circle, passing through the points $(0,0),(1,0)$ and touching the circle $x^2+y^2=9$, be…

Solution