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(h^2+k^2) is equal to __________.

Circle will touch internally\ C_1 C_2=|r_1-r_2| \ =sqrt(h^2+k^2)=3-sqrt(h^2+k^2) \ => 2 sqrt(h^2+k^2)=3 \ => h^2+k^2=(9 / 4) \ Therefore 4(h^2+k^2)=9 \

Let the centre of a circle passing through the points 0... | (Mathematics)

Circle will touch internally

$\\ \begin{aligned} & C_1 C_2=\left|r_1-r_2\right| \\ & =\sqrt{h^2+k^2}=3-\sqrt{h^2+k^2} \\ & \Rightarrow 2 \sqrt{h^2+k^2}=3 \\ & \Rightarrow h^2+k^2=\frac{9}{4} \\ & \therefore 4\left(h^2+k^2\right)=9 \end{aligned}$

Explanation

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