Find area of right triangle △ with circumcircle $O$ and tangent circle $K$.

Question

The triangle $ABC$, with $\angle A = 90^\circ$, is inscribed in the circle $\mathcal{O}$. The circle $\mathcal{K}$ is tangent to the circle $\mathcal{O}$, the sides $AB$ and $AC$. Suppose that the radius of $\mathcal{O}$ is $3$ and the radius of $\mathcal{K}$ is $2$. Find the area of $\triangle ABC$.

I drew this problem out by making circle $\mathcal{K}$ internally tangent to circle $\mathcal{O}$. I was able to make it tangent to $AB$ and $AC$. However, I have gotten nowhere using trigonometry to solve this.

I also tried to name the sides and angles to find the area, which did not work, either

Answer 1

One possible solution is to assume a special case for which the second circle information is superfluous.

For example, inscribe a square into circle O with a corner at point A. Then, per a source, to quote:

We can show this using a symmetry argument – the square is symmetrical across its diagonal, so the diagonal must pass through the center of the circle.

The diagonal line also creates two right triangles, one at Point A. So, with a hypotenuse of 6 (as the radius is given as 3), the sides of the triangle (and square) are ${3\sqrt{2}}$.

This implies an area of the square of 18 sq units and for the triangle, as one possible answer for the question, 9 sq units.

Answer 2

Start with arbitrary point $L$ on the circle $\mathcal{O}(O,R)$ and draw the circle $\mathcal{K}(K,r_k)$, tangent to $\mathcal{O}(O,R)$ at $L$.

Let $L=(0,-R)$, $K=(0,-R+r_k)$.

Since $\mathcal{K}$ is inscribed in $\angle CAB$, and $\angle CAB=90^\circ$, the distance $|KA|=r_k\sqrt{2}$, so we can find the point $A$ as the intersection of the $\mathcal{O}$ and the circle centered at $K$ with the radius $r_k\sqrt2$.

$AK$ is the bisector of $\angle CAB$, hence the tangent point $D= AB\cap \mathcal{K}$ can be easily found, as $\triangle ADK$ is isosceles.

When $R=3$, $r_k=2$, $\triangle OKA$ is right angled, since in this case $|OK|^2+|KA|^2=|OA|^2=R^2=9$, which means that

\begin{align} |AD|=|KD|&=r_k=2 ,\\ |OE|&=\frac{r_k}{\sqrt2}=\sqrt2 ,\\ |DE|&=\frac{r_k}{\sqrt2}-(R-r_k)=\sqrt2-1 ,\\ |OD|&= \sqrt{R^2-(2+\sqrt2)\,r_k\,(R-r_k)} =\sqrt{5-2\sqrt2} . \end{align}

Then consider $\triangle ABO$, for which we already know values of $AD$, $|OD|$ and also that $|OB|=|OA|=3$, so we can apply the Stewart’s theorem

\begin{align} \triangle ABO:\quad &|OB|^2\cdot|AD|+|OA|^2 \cdot(|AB|-|AD|) \\ &= |AB|\cdot(|OD|^2+|AD|\cdot(|AB|-|AD|)) , \end{align}

which gives the unique solution: $|AB|=4+\sqrt2$. Then \begin{align} |AC|&=\sqrt{4R^2-|AB|^2} =4-\sqrt2 \end{align}

and the sought area then is

\begin{align} S_{ABC}&=\tfrac12\,|AB|\cdot|AC|= \tfrac12(16-2)=7 . \end{align}

Answer 3

Note that BK $=2\sqrt2$, the diagonal of the square of side length $2$ and $\triangle$BOK is a right triangle because of its sides, $1$, $2\sqrt2$ and $3$, satisfying the Pythagorean rule. Then, $\sin\theta =\frac{KO}{BO}= \frac13$ and $$\sin\alpha \cos\alpha = \sin(45-\theta)\cos(45-\theta)=\frac{1}2\cos2\theta=\frac12-\sin^2\theta=\frac7{18} $$ Thus, the area of $\triangle$ABC is $$Area_{ABC}=\frac12AB\cdot BC=\frac12AC^2\sin\alpha\cos\alpha = \frac12\cdot6^2\cdot\frac7{18}=7 $$