Question

Let ABCD be a square of sides of sides of length $2$ units. $C_1$ is the circle touching all the sides of the square and $C_2$ is the circle circumscribing the square. $l$ is a line through A.
(i) If P is a point on $C_1$ and Q is a point on $C_2$ then $\frac{P{A}^2+P{B}^2+P{C}^2+P{D}^2}{Q{A}^2+Q{B}^2+Q{C}^2+Q{D}^2}$ equals

• A. $\frac{3}{4}$
• B. $\frac{5}{4}$
• C. $1$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $\frac{3}{4}$

Let the, equation of circle are
$C_1:{(x-1)}^2+{(y-1)}^2={\left(1\right)}^2$
and $C_2:{(x-1)}^2+{(y-1)}^2={\left(\sqrt{2}\right)}^2$
Therefore coordinate of $P(1+\cos \theta ,1+\sin \theta )$ and $Q(1+\sqrt{2}\cos \theta ,1+\sqrt{2}\sin \theta )$
$\therefore P{A}^2+P{B}^2+P{C}^2+P{D}^2=\{{(1+\cos \theta )}^2+{(1+\sin \theta )}^2\}+\{{(\cos \theta -1)}^2+{(1+\sin \theta )}^2\}+\{{(\cos \theta -1)}^2+{(\sin \theta -1)}^2\}+\{{(1+\cos \theta )}^2+{(\sin \theta -1)}^2\}=12$
Similarly $Q{A}^2+Q{B}^2+Q{C}^2+Q{D}^2=16$
$\therefore \frac{\sum P{A}^2}{\sum Q{A}^2}=\frac{12}{16}=0.75$