Finding the area of a triangle in a circle from a point.

Question

$P$ is variable point. Tangents are drawn to the circle $x^2+y^2=4$ from $P$ to touch it at $ Q$ and $R$. The parallelogram $PQSR$ is completed. If $P $ is $(6,8) $, then find the area of $QSR$.

I have been trying to solve this question since a long time but I'm unable to do so. I tried to find the coordinates of intersection of the tangents but the calculation was a mess. I also tried to use the fact that $PQOR$ is a cyclic quadrilateral but it did not prove to be helpful. I have run out of ideas. Would someone please help me to solve this question in an elegant and simple manner?

Answer 1

The distance from $P$ to the origin is $\sqrt{6^2+8^2} = 10$, so it is equivalent to let the point $P$ at (0,10) due to rotation invariance around the origin and the computation gets a lot easier due to symmetry.

Let $2\theta = \angle QPR$. Then, you have

$$\sin\theta = \frac{2}{10}= \frac{1}{5}$$

Since $\angle RQO = \theta$ due to similar triangles, then the area QSR is

$$ (10-2\sin\theta)\cdot2\cos\theta$$

Answer 2

Since $PQSR$ is a paralleogram, then area of $\triangle QSR$ is the same as the area of $\triangle QPR$.

Since $PQ$ and $PR$ are fragments of lines tanget to the circle, it means that traingles $PQ \perp QO$ and $PR \perp RO$. It also means that $QR \perp PO$.

Let $K$ be the intersection point of lines $PO$ and $RQ$. We have $$ {\rm Area}(\triangle PQR) = 2 \cdot {\rm Area}(\triangle PQK)$$ Then triangles $PKQ$ and $PQO$ are similar right triangles, with the scale of similiarity being $$ \lambda = {{|PQ|}\over{|PO|}}$$ That means that $$ {\rm Area}(\triangle PQK) = \lambda^2 \cdot {\rm Area}(\triangle PQO) = \frac{|PQ|^2}{|PO|^2} \cdot \frac12 |PQ| |QO| = \frac{|PQ|^3|QO|}{2|PO|^2}$$ $$ {\rm Area}(\triangle PQR) = \frac{|PQ|^3|QO|}{|PO|^2}$$ We have $|PO| = \sqrt{6^2+8^2} = 10$, $|QO|=2$ (the radius of the circle), and from the Pythagoras' theorem, $|PQ|=\sqrt{10^2-2^2}=4\sqrt{6}$. That means that $$ {\rm Area}(\triangle PQR) = \frac{(4\sqrt{6})^3\cdot 2}{10^2} = \frac{192\sqrt{6}}{25}$$

Answer 3

Let $Q = (x_0, y_0)$ And let $R=(x_1, y_1)$.

So $x_0^2 + y_0^2 = 4$. The line from the origin of the circle $O$ to $Q$ is $\frac {y_0}{x_0}$. So the slope of the tangent line is $-\frac {x_0}{y_0}$.

$P =(6,8)$ is on this line so $y_0 - 8 = -(x_0 -6)\frac {x_0}{y_0}$

Or $y_0^2 - 8y_0 = -x_0^2 + 6x_0$

Or $x_0^2 + y_0^2 = 6x_0 + 8y_0$

Or $4 = 6x_0 + 8y_0$

Or $y_0 = -\frac 34 x_0+\frac 12$.

But we know $x_0^2 + y_0^2 = 4$ so $x_0^2 + (-\frac 34 x_0+\frac 12)^2 = 4$.

Some thing can be done for $x_1, y_1$

We will have $y_1 =-\frac 34 x_1 + \frac 12$ and $x_1^2 + (-\frac 34 x_1+\frac 12)^2 = 4$.

So...

Solve the equation $x^2 + (-\frac 34x + \frac 12)^2 = 4$.

You will get two answers. One will be $x_0$ and the other will by $x_1$. Plug them into the equations $y_i = -\frac 34 x_i+\frac 12$ to find $y_0$ and $y_1$.

Those we know the coordinates of $P,Q, R$. $S$ can be found by completing the parallelogram but may not be needed.

Can you finish?