JEE MAIN - Mathematics (2024 - 9th April Morning Shift - No. 26)
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 __________.
Answer
9
Explanation
Circle will touch internally
$$\begin{aligned} & C_1 C_2=\left|r_1-r_2\right| \\ & =\sqrt{h^2+k^2}=3-\sqrt{h^2+k^2} \\ & \Rightarrow 2 \sqrt{h^2+k^2}=3 \\ & \Rightarrow h^2+k^2=\frac{9}{4} \\ & \therefore 4\left(h^2+k^2\right)=9 \end{aligned}$$
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