# 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 $(a^2+b^2+ab)$ is equal to ________ Option: 1 9 Option: 2 10 Option: 3 5 Option: 4 7

All normals of a circle pass through center Radius = CA = CB

$\\\mathrm{CA^2=CB^2}\\ (a-3)^{2}+(b+3)^{2} =(a-4)^{2}+(b-2 \sqrt{2})^{2} \\ a+(3-2 \sqrt{2}) b=3 \\ a-2 \sqrt{2} b+3 b=3 \\ \text { given that } a-2 \sqrt{2} b=3\\ \text{from above, we get }a=3 \text{ and }b=0\\a^{2}+b^{2}+a b=9$

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