Let P(a_1,b_1) and Q(a_2,b_2) be two distinct points on a circle with center C(√2,√3).

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 the origin and $OC$ be perpendicular to both $CP$ and $CQ$. If the area of the triangle $OCP$ is $\frac{\sqrt{35}}{2}$, then ${a}_{2}^{1}+{a}_{2}^{2}+{b}_{2}^{1}+{b}_{2}^{2}$ is equal to :

Answer: 24

Accepted Answer

$OC\, \perp\, CP$ and $OC\, \perp\, CQ$

$\Rightarrow PCQ$ is a straight line

$OC=\sqrt{{(\sqrt{2})}^{2}+{(\sqrt{3})}^{2}}=\sqrt{5}$

Let $CP=CQ=I$

$[OCP]=\frac{1}{2}	imes OC	imes I=\frac{\sqrt{35}}{2}$

$I=\sqrt{7}$

$OP=OQ=\sqrt{{(OC)}^{2}+{I}^{2}}=\sqrt{5+7}=\sqrt{12}$

${a}_{2}^{1}+{a}_{2}^{2}+{b}_{2}^{1}+{b}_{2}^{2}=({a}_{2}^{1}+{b}_{2}^{2})+({a}_{2}^{2}+{b}_{2}^{2})$

$O{P}^{2}+O{Q}^{2}=12+12=24$

Solutions · 1