Let normals at all points on a given curve pass through a fixed point (a, b) . If curve passes throu

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10-1.Circle and System of Circles

hard

Let the normals at all the points on a given curve pass through a fixed point $(a, b) .$ If the curve passes through $(3,-3)$ and $(4,-2 \sqrt{2}),$ and given that $a-2 \sqrt{2} b=3,$ then $\left(a^{2}+b^{2}+a b\right)$ is equal to ..... .

A

$6$

B

$3$

C

$4$

D

$9$

(JEE MAIN-2021)

Solution

All normals of circle passes through centre Radius $= CA = CB$

$CA ^{2}= CB ^{2}$

$( a -3)^{2}+( b +3)^{2}$

$=( a -4)^{2}+( b -2 \sqrt{2})^{2}$

$a+(3-2 \sqrt{2}) b=3$

$a-2 \sqrt{2} b+3 b=3 ….(1)$

given that $a -2 \sqrt{2} b =3 ….(2)$

from $(1) \&(2) \Rightarrow a=3, b=0$

$a^{2}+b^{2}+a b=9$

Standard 11

Mathematics

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