Maximum area of $\Delta QSR$

Question

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

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

$$A(\theta)=\sqrt{2}sin^2\theta(\sin\theta+\cos\theta)$$

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

Answer 1

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 $\theta.$

Answer 2

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.