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.

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 $7$.

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.