Let lines y+2 x=√{11}+7 √{7} and 2 y + x =2 √{11}+6 √{7} be normal to a circle C:(x-h)^{2}+(

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10-1.Circle and System of Circles

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Let the lines $y+2 x=\sqrt{11}+7 \sqrt{7}$ and $2 y + x =2 \sqrt{11}+6 \sqrt{7}$ be normal to a circle $C:(x-h)^{2}+(y-k)^{2}=r^{2}$. If the line $\sqrt{11} y -3 x =\frac{5 \sqrt{77}}{3}+11$ is tangent to the circle $C$, then the value of $(5 h-8 k)^{2}+5 r^{2}$ is equal to.......

A

$916$

B

$816$

C

$856$

D

$86$

(JEE MAIN-2022)

Solution

Normal are

$y +2 x =\sqrt{11}+7 \sqrt{7}$

$2 y + x =2 \sqrt{11}+6 \sqrt{7}$

Center of the circle is point of intersection of ormals i.e.

$\left(\frac{8 \sqrt{7}}{3}, \sqrt{11}+\frac{5 \sqrt{7}}{3}\right)$

Tangent is $\sqrt{11} y-3 x=\frac{5 \sqrt{77}}{3}+11$

Radius will be $\perp$ distance of tangent from center

i.e. $4 \sqrt{\frac{7}{5}}$

Now $(5 h -8 k )^{2}+5 r ^{2}=816$

Standard 11

Mathematics

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