JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
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 \(9\)
Answer & Solution
Correct Answer
(D) \(9\)
Step-by-step Solution
Detailed explanation
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…
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