Let the maximum and minimum values of (sqrt 8 x-x 2 -12 -4 ) 2 +(x-7) 2 , x in R be M and m respectively. Then M 2 -m 2 is equal to

Mathematics · Circles

JEE Main 2024 — 5 April, Shift 2 — Question 28

Let the maximum and minimum values of $\left(\sqrt{8 x-x^{2}-12}-4\right)^{2}+(x-7)^{2}, x \in R$ be $M$ and $m$ respectively.

Then $M^{2}-m^{2}$ is equal to $\qquad$

Numerical answer — enter this value.

$(x-7)^{2}+(y-4)^{2}$

$y=\sqrt{8 x-x^{2}-12}$

$y^{2}=-(x-4)^{2}+16-12$

$(x-4)^{2}+y^{2}=4$

$\mathrm{m}=9$

$\mathrm{M}=41$

$M^{2}-m^{2}=41^{2}-9^{2}=1600$

Answer key and solution verified before publishing.

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