The circle $C \equiv x^2+y^2=1$ cuts $X$ and $Y$ axes at $P$ and $Q$ Respectively. if another circle with centre $Q$ and variable radius is drawn so that it meets $C$ at $R$ and the line $PQ$ at $S$. Find the Maximum area of $\Delta QSR$

My Try: Let $Q(0,1)$ and $P(1,0)$ Let the radius of variable circle be $r$ so its equation is $$C' \equiv x^2+(y-1)^2=r^2$$.

Let the Point $R$ be $R(r\cos\theta,\: r\sin\theta+1)$ .Now since $R$ also lies on $C$ we have

$$(r\cos\theta)^2+(r\sin\theta+1)^2=1$$ $\implies$

$$r=-2\sin\theta$$ Hence $R$ is $R(-\sin2\theta, \cos2\theta)$

Since $S$ is a point on both the circle $C'$ and the line $PQ \equiv x+y-1=0$, its coordinates are

$S (\frac{r}{\sqrt{2}}, 1-\frac{r}{\sqrt{2}})=(-\sqrt{2}\sin\theta, 1+\sqrt{2}\sin\theta)$

Now area of $\Delta QSR$ is given by absolute value of

$$A(\theta)=\frac{1}{2}\begin{vmatrix} 0 & 1 &1 \\ -\sin2\theta & \cos2\theta & 1\\ -\sqrt{2}\sin\theta & 1+\sqrt{2}\sin\theta& 1 \end{vmatrix}=\begin{vmatrix} 0 & 1 &0 \\ -\sin2\theta & \cos2\theta & 2\sin^2\theta\\ -\sqrt{2}\sin\theta & 1+\sqrt{2}\sin\theta& -\sqrt{2}\sin\theta \end{vmatrix} $$


Now to find Maximum we need to differentiate and find $\theta$. Can i have any better approaches?


enter image description here

As seen in the figure, there are 4 possible triangles meeting your description of ⊿QRS. We will only consider the case of the one in RED.

It would be easier if we define $\angle POR = \theta$ as shown. Then, $R = (\cos \theta, \sin \theta)$ and $\angle RQS = \dfrac {\theta}{2}$.

R is also a point on C’. This will give $r^2 = 2 – 2\sin \theta$

$[\triangle QRS] = \dfrac { r^2 \sin \dfrac {\theta }{2}}{2} = (1 – \sin \theta)( \sin \dfrac {\theta }{2})$

Differentiating the above will yield the optimal $\theta$ that makes the required area maximum.


There may be a sign error and a missing factor $1/2$ in front of the intermediate result but the general approach is sound. Be careful in the problem statement because the way you formulate it, $P$ and $Q$ would not be unique. If you need explicit proof that it is a maximum you could either take the second derivative or evaluate $A$ for specific choices of $\theta.$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.