JEE MAIN - Mathematics (2024 - 5th April Evening Shift - No. 23)
Let the maximum and minimum values of $$\left(\sqrt{8 x-x^2-12}-4\right)^2+(x-7)^2, x \in \mathbf{R}$$ be $$\mathrm{M}$$ and $$\mathrm{m}$$, respectively. Then $$\mathrm{M}^2-\mathrm{m}^2$$ is equal to _________.
Answer
1600
Explanation
$$\begin{aligned} & \text { Let } y=\sqrt{8 x-x^2-12} \Rightarrow(x-4)^2+y^2=2^2 \\ & \Rightarrow d=(y-4)^2+(x-7)^2 \end{aligned}$$
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$$\begin{aligned} \Rightarrow & M=P A^2=16+25=41 \\ & m=P Q^2=(\sqrt{16+9}-2)^2=9 \\ \Rightarrow & M^2-m^2=1681-81=1600 \end{aligned}$$
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