2 . Passage for Q. (395 to 397) : Let $A B C D$ be a square of side length 2 units. $C_{2}$ is the c

Question

2 . Passage for Q. (395 to 397) : Let $A B C D$ be a square of side length 2 units. $C_{2}$ is the circle through vertices $A, B, C, D$ and $C_{1}$ is the circle touching all the sides of the square $A B C D$. $L$ is a line through $A$.\If $P$ is a point on $C_{1}$ and $Q$ is another point on $C_{2}$, then $\frac{P A^{2}+P B^{2}+P C^{2}+P D^{2}}{Q A^{2}+Q B^{2}+Q C^{2}+Q D^{2}}$ is equal to\

Professor PC Thomas Classes · Answer

(a) : Let $A, B, C$ and $D$ be the complex numbers\ $\sqrt{2},-\sqrt{2}, \sqrt{2} i$ and $-\sqrt{2} i$ respectively\ $\Rightarrow \frac{P A^{2}+P B^{2}+P C^{2}+P D^{2}}{Q A^{2}+Q B^{2}+Q C^{2}+Q D^{2}}$\ $=\frac{\left|z_{1}-\sqrt{2}ight|^{2}+\left|z_{1}+\sqrt{2}ight|^{2}+\left|z_{1}+\sqrt{2} iight|^{2}+\left|z_{1}-\sqrt{2} iight|^{2}}{\left|z_{2}+\sqrt{2}ight|^{2}+\left|z_{2}-\sqrt{2}ight|^{2}+\left|z_{2}-\sqrt{2} iight|^{2}+\left|z_{2}+\sqrt{2} iight|^{2}}$\ $=\frac{\left|z_{1}ight|^{2}+2}{\left|z_{2}ight|^{2}+2}=\frac{3}{4}$.\