Question

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 \((4,-2\sqrt2)\), and given that \(a-2\sqrt2\,b=3\), then \((a^2+b^2+ab)\) is equal to __________.

Hint:

If all normals pass through one point, the curve is a circle with that point as centre; equate the distances of the two given points from \((a,b)\).

Answer 1

Correct Answer: 9

All normals pass through \((a,b)\) only for a circle centred at \((a,b)\). Hence \((3-a)^2+(-3-b)^2=(4-a)^2+(-2\sqrt2-b)^2\).
\(18-6a+6b+a^2+b^2=24-8a+4\sqrt2\,b+a^2+b^2\Rightarrow2a+6b-4\sqrt2\,b=6\Rightarrow a+3b-2\sqrt2\,b=3\).
Given \(a-2\sqrt2\,b=3\), subtracting gives \(3b=0\Rightarrow b=0\), \(a=3\).
\(a^2+b^2+ab=9+0+0=9\).
Answer: 9