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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 \mathrm{(4,-2 \sqrt{2})}, and given that \mathrm{a-2 \sqrt{2} b=3}, then \mathrm{\left(a^2+b^2+a b\right)} is equal to __________.

Option: 1

8


Option: 2

9


Option: 3

10


Option: 4

7


Answers (1)

best_answer

All normals of circle pass through a fixed point that is the centre.

\mathrm{ \therefore \text { Radius }=C A=C B }

\mathrm{ \begin{aligned} \Rightarrow & C A^2=C B^2 \\\\ \Rightarrow & (a-3)^2+(b+3)^2 \\\\ & =(a-4)^2+(b+2 \sqrt{2})^2 \\\\ \Rightarrow & -6 a+8 a+6 b-4 \sqrt{2} b=6 \\\\ \Rightarrow & a-2 \sqrt{2} b+3 b=3 \end{aligned} }             ...(i)

Also, given that \mathrm{ a-2 \sqrt{2} b=3}                          ...(ii)

From (i) and (ii), we get, \mathrm{a=3, b=0}

\mathrm{ \therefore a^2+b^2+a b=9 }

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