JEE Main 2024 — 9 April, Shift 1 — Question 28

$(x-h)^{2}+(y-k)^{2}=h^{2}+k^{2}$

$x^{2}+y^{2}-2 h x-2 k y=0$

$\because$ passes through $(1,0)$

$\Rightarrow 1+0-2 \mathrm{~h}=0$

$\Rightarrow \mathrm{h}=1 / 2$

$\because \mathrm{OC}=\frac{\mathrm{OP}}{2}$

$\sqrt{\left(\frac{1}{2}\right)^{2}+\mathrm{k}^{2}}=\frac{3}{2}$

$\frac{1}{4}+\mathrm{k}^{2}=\frac{9}{4}$

$\mathrm{k}^{2}=2$

$\mathrm{k}= \pm \sqrt{2}$

$\therefore$ Possible coordinate of $\mathrm{c}(\mathrm{h}, \mathrm{k})\left(\frac{1}{2}, \sqrt{2}\right)\left(\frac{1}{2},-\sqrt{2}\right)$

$4\left(\mathrm{~h}^{2}+\mathrm{k}^{2}\right)=4\left(\frac{1}{4}+2\right)=4\left(\frac{9}{4}\right)=9$