Find the number of possible points $R$. 
$P(3,1),Q(6,5)$ and $R(x,y)$ are three points such that the angle $\angle PRQ=90^{\circ}$ and the area of the triangle $\triangle PRQ=7$.The number of such points $R$ that are possible is .

$a.)\ 1\\
b.)\ 2\\
c.)\ 3\\
d.)\ 4\\
e.)\ 0\\$
For $\angle PRQ$ to be right angled, The point should lie on equation of circle with $PR$ as diameter.
$\left(x-\dfrac92\right)^2+(y-3)^2=\left(\dfrac52\right)^2$
And with the formula for area of triangle i found,
$4x-3y=-5\ \text{or}\\
4x-3y=23$

By graphing it i found that no possible values for $R$.
How do i find that by pen and paper.
I look for a short and simple method.
I have studied maths upto $12th$ grade.
 A: We know that $R$ has to be on the circle whose diameter is $PQ(=5)$. Then, note that for a point $S$ on the circle, the maximum of the area of $\triangle{PQS}$ is $PQ\times\frac{PQ}{2}\times \frac 12=\frac{25}{4}$. This is smaller than $7$. 
A: We define the vectors
$$
u = RP = P - R = (3,1) - (x, y) = (3-x, 1-y) \\
v = RQ = Q - R = (6,5) - (x, y) = (6-x, 5-y)
$$
From the angle condition we know $u$ and $v$ are orthogonal, thus
$$
u \cdot v = 0
$$
The area of the formed triangle is the half of the formed rectangle:
$$
7 = \frac{1}{2} \lVert u \rVert \lVert v \rVert
$$
This gives the system
$$
0 = (3-x)(6-x) + (1-y)(5-y) \\
14 = \sqrt{(3-x)^2 + (1-y)^2} \sqrt{(6-x)^2 + (5-y)^2}
$$
The first equation can be rewritten to
$$
0 = 18 + 5 + x^2 + y^2 - 9 x - 6 y = 23 + (x - 4.5)^2 + (y-3)^2 - 81/4 - 9 \Rightarrow \\
(x - 4.5)^2 + (y - 3)^2 = (5/2)^2
$$
so this is a Thales circle with radius $r=5/2$ and center $C=(4.5,3)$.
The other equation is some 4th order curve. Plotting it gives

And one notes there is no intersection, thus no solution.

For a Thales circle $x^2 + y^2 = 1$ we have the area
$$
u = RP = P - R = (-1-x,-y) \\
v = RQ = Q - R = (1 - x, -y)
$$
and
\begin{align}
A(x,y) 
&= \frac{1}{2} \sqrt{(1+x)^2 + y^2} \sqrt{(1-x)^2 + y^2} \\
&= \frac{1}{2} \sqrt{(1+x)^2 + 1 - x^2} \sqrt{(1-x)^2 + 1 - x^2} \\
&= \frac{1}{2} \sqrt{2 + 2x} \sqrt{2 - 2x} \\
&= \sqrt{1 - x^2} \\
&= \sqrt{y^2} \\
&= \lvert y \rvert
\end{align}
which is maximal for $(x, y) = (0, \pm 1)$ and value $A = 1$.

Thus a Thales circle with radius $r$ has maximal area $r^2$.
This means for the original Thales circle the area is less equal $r^2 = 25/4 < 28/4 = 7$, so there is never enough area to solve the problem.
