Coordinate Geometry
Circles
MMTS_Full_Test_05
Grade 12
Question:
Let $P(a_1,b_1)$ and $Q(a_2,b_2)$ be two distinct points on a circle with center $C(\sqrt{2},\sqrt{3})$. Let $O$ be origin and $OC$ be perpendicular to both $CP$ and $CQ$. If the area of $\triangle OCP$ is $\sqrt{35}$, then $a_1^2+a_2^2+b_1^2+b_2^2$ is equal to
Step-by-Step Solution
Key Concept: $OC\perp CP$ and $OC\perp CQ$ means $P$ and $Q$ are symmetric about the line $OC$
$|OC|=\sqrt{5}$. Area$=\frac{1}{2}\sqrt{5}\cdot|CP|=\sqrt{35}\Rightarrow|CP|=2\sqrt{7}$. $P$ on circle: $|CP|=r=2\sqrt{7}$. $a_1^2+b_1^2=|OP|^2=|OC|^2+|CP|^2=5+28=33$ (wait: $OP^2=OC^2+CP^2$ only if right angle at $C$). $a_1^2+b_1^2+a_2^2+b_2^2=2\cdot(5+28)=66$.
Correct Answer: 66