The question is as follows:
Let $(a_1,b_1)$ and $Q(a_2,b_2)$ be two distinct points on a circle with center $C(√2 , √3)$. Let $O$ be the origin and $OC$ be perpendicular to both $CP$ and $CQ$. If the area of the triangle $OCP$ is $\sqrt 35 /2$ ,then $a_1^2 + a_2^2 + b_2^2 + b_2^2$ is equal to _____.
And here is my solution:
Since OC - OP, and OC - OQ meet perpendicular to each other, PQ must form a straight line. Since C is the center, $\Delta OCP = \Delta OCQ$. Using distance formula, we can say that: $$PC = \sqrt 5$$
Since OPQ form a triangle of area $\sqrt 35$,we can say that: $$\sqrt 35 = \frac{1}{2} \sqrt 5 \sqrt {(a_1^2 - a_2^2) + (b_1^2 - b_2^2)}$$ $$28 = (a_1^2 - a_2^2) + (b_1^2 - b_2^2)$$ $$ 28 = a_1^2 + a_2^2 + b_1^2 + b_2^2 - 2(a_1a_2+b_1b_2)$$
But the answer turns out to be $a_1^2 + a_2^2 + b_1^2 + b_2^2 = 24$. So it must be sufficient for me to prove that $a_1a_2 + b_1b_2$ is equal 2. But don't see how I can. Any help is appreciated!