I have this circle problem :
I think this question requires some more data which is not provided. Here is what I have logically tried so far :
For the circle with center O, let's say the radius is $r_1$. So OP = OQ = $r_1$. From the figure, it's apparent that $\angle POQ = {90}^{\circ}$. Hence, PQ = $r_1 * \sqrt 2$. Now for $\triangle OPC$, $\angle OPC = {60}^{\circ} $, $\angle PCO = {30}^{\circ}$ and OP : OC : PC = 1 : $\sqrt 3$ : 2. So, CQ = OC - OQ = $r_1 * \sqrt 3 - r_1$. Now area of $\triangle PQC$ is $\frac{1}{2} * CQ * OP = \frac{1}{2} * (r_1 * \sqrt 3 - r_1) * r_1 = \frac{\sqrt 3 - 1}{2} * {r_1}^{2}$. But there is no exact mention of the value $r_1$.
There is no explanation given for the right answer marked and I can't wrap my head around how they have reached this answer. Am I missing something very obvious? Is there anything wrong in my logic? Any help would be helpful.
Your work is correct. The diagram is badly drawn as $CQ$ should be about $\frac 34$ of $OQ$. You are correct that without something to set the scale the question cannot be answered. Your answer assumes $r_1=1$, which seems reasonable to me but is not specified.