JEE Mathematics 2023 Question with Solution

Given: $C(\sqrt{2}, \sqrt{3})$ is the center of the circle, so

Also, $OC \perp CP$ and the area of triangle $OPC$ is

Find: $a_1^2 + a_2^2 + b_1^2 + b_2^2$.

From

we get

Since $P$ and $Q$ lie on the same circle centered at $C$,

Now,

Using the right triangle relation shown in the working,

and similarly,

Therefore,

So, the required value is $24$.

Given: $OC$ is perpendicular to both $CP$ and $CQ$, with $OC = \sqrt{5}$ and area of triangle $OPC$ equal to $\frac{\sqrt{35}}{2}$.

Find: The value of $OP^2 + OQ^2$.

Because $C(\sqrt{2}, \sqrt{3})$,

Hence,

The area condition gives

so

and therefore

Since $P$ and $Q$ are on the same circle with center $C$,

Now triangle $OCP$ is right-angled at $C$, so

Similarly,

Thus,

Therefore, the correct answer is $24$.