Kite Problem!

Let $AC=BC=l$ . We note that $\angle CBO = \angle CAO = 90^\circ$ . Then $A_{OACB} = 2 \times \dfrac {rl}2 = rl = \sqrt{\dfrac {11}5}r^2 \implies l = \sqrt{\dfrac {11}5} r \implies \dfrac l r = \sqrt{\dfrac {11}5}$ . We also note that $\triangle BMO$ and $\triangle CBO$ are similar. Therefore $\dfrac {BM}{OM} = \dfrac lr = \sqrt{\dfrac {11}5} \implies BM = \sqrt{\dfrac {11}5} OM = \sqrt{\dfrac {11}5} \times \sqrt{(1-0)^2+(2-0)^2} = \sqrt{11}$ . By Pythagorean theorem , $r^2 = OM^2 + BM^2 = 5 + 11 = 16 \implies r = \boxed 4$ .

$OM = \sqrt{5} \implies MB = MA = \sqrt{r^2 - 5} \implies AB = 2\sqrt{r^2 - 5} \implies$

$A_{\triangle{OAB}} = \dfrac{1}{2}(AB)(OM) = \sqrt{5}\sqrt{r^2 - 5}$

$\triangle{OAM} \sim \triangle{MAC} \implies$ $\dfrac{\sqrt{r^2 - 5}}{\sqrt{5}} = \dfrac{h}{\sqrt{r^2 - 5}} \implies h = \dfrac{r^2 - 5}{\sqrt{5}} \implies$

$A_{\triangle{ACB}} = \dfrac{1}{2}(AB)(h) = (r^2 - 5)\dfrac{\sqrt{r^2 - 5}}{\sqrt{5}}$

$\implies A_{OACB} = A_{\triangle{OAB}} + A_{\triangle{ACB}} = \sqrt{\dfrac{r^2 - 5}{5}} r^2 =$ $\sqrt{\dfrac{11}{5}}r^2$

$\implies \dfrac{r^2 - 5}{5} = \dfrac{11}{5} \implies r = \boxed{4}$

The equation of the line $OM$ is $y=2x$ . Hence, ${{y}_{C}}=2{{x}_{C}}$ .
$M$ is the midpoint of $AB$ , thus $OM\bot AB$ , so the gradient of the line $AB$ is $-\dfrac{1}{2}$
and its equation is $y-2=-\dfrac{1}{2}\left( x-1 \right)\Leftrightarrow y=-\dfrac{1}{2}x+\dfrac{5}{2}\ \ \ \ \ \left( 1 \right)$ .

On the other hand, $AB$ is the polar of the point $C$ with respect to the circle, therefore its equation can be written as:
${{x}_{C}}\cdot x+{{y}_{C}}\cdot y={{r}^{2}}$ which rearranges to $y=-\frac{{{x}_{C}}}{{{y}_{C}}}x+\dfrac{{{r}^{2}}}{{{y}_{C}}}\Leftrightarrow y=-\dfrac{1}{2}x+\dfrac{{{r}^{2}}}{2{{x}_{C}}} \ \ \ \ \ \left( 2 \right)$

$\left( 1 \right),\left( 2 \right)\Rightarrow \dfrac{{{r}^{2}}}{2{{x}_{C}}}=\dfrac{5}{2}\Leftrightarrow {{x}_{C}}=\dfrac{{{r}^{2}}}{5}$ , therefore, the coordinates of the point $C$ are $\left( \dfrac{{{r}^{2}}}{5},\ \dfrac{2{{r}^{2}}}{5} \right)$ .
Hence, $OC=\sqrt{{{x}_{C}}^{2}+{{y}_{C}}^{2}}=\dfrac{{{r}^{2}}}{\sqrt{5}}$ and $AC=\sqrt{\dfrac{{{r}^{4}}}{5}-{{r}^{2}}}=\sqrt{\dfrac{{{r}^{4}}}{5}-{{r}^{2}}}$ .

This gives ${{A}_{OACB}}= 2{{A}_{OAC}}=2\cdot \frac{1}{2}\cdot OA\cdot AC=r\sqrt{\frac{{{r}^{4}}}{5}-{{r}^{2}}}$ .

Finally, $\begin{aligned} {{A}_{OACB}}=\sqrt{\frac{11}{5}}r & \Leftrightarrow r\sqrt{\frac{{{r}^{4}}}{5}-{{r}^{2}}}=\sqrt{\frac{11}{5}}r \\ & \Leftrightarrow {{r}^{2}}\left( \frac{{{r}^{4}}}{5}-{{r}^{2}} \right)=\frac{11}{5}{{r}^{2}} \\ & \Leftrightarrow r=\boxed{4}. \\ \end{aligned}$

Let $\overline {OB}$ and $\overline {OA}$ make angles $α$ and $β$ with the $x$ -axis. Then slope of $\overline {AB}$ is

$\dfrac {\sin α-\sin β}{\cos α-\cos β}=-\cot (\frac{α+β}{2})$

Slope of $\overline {OC}$ is $2$ . Since $\overline {AB}$ is perpendicular to $\overline {OC}$ , therefore

$\tan (\frac{α+β}{2})=2\implies \cos (\frac{α+β}{2})=\dfrac {1}{\sqrt 5}$

$|\overline {AC}|=r\tan (\frac{β-α}{2})$

Areas of triangles $\triangle {OAC}$ and $\triangle {OBC}$ are equal. So, area of $OACB$ is

$2\times \dfrac 12\times r\times r\tan (\frac{β-α}{2})=r^2\tan (\frac{β-α}{2})$

$=\sqrt {\dfrac {11}{5}}r^2$

$\implies \tan (\frac{β-α}{2})=\sqrt {\dfrac {11}{5}}$

$\implies \cos (\frac{β-α}{2})=\dfrac {\sqrt 5}{4}$

Also, we have

$\dfrac {r\cos α+r\cos β}{2}=1$

$\implies r=\dfrac {2}{\cos α+\cos β}$

$=\dfrac {1}{\cos (\frac{α+β}{2})\cos (\frac{β-α}{2})}$

$=\dfrac {1}{\frac{1}{\sqrt 5}\times \frac{\sqrt 5}{4}}=\boxed 4$ .

Rotation about the origin will preserve the distance $\left | OM\right |\mapsto\left | OM'\right |$ . We then imagine the kite rotated around such that $C$ is above the origin, as shown. https://www.desmos.com/calculator/hysbdxaoc9

$\left | OM'\right |=\sqrt{2^2+1^2}\implies M'=\left (0, \sqrt 5\right )$ . Since $OA'$ and $OB'$ are radii, then $A'=\left (-\sqrt{r^2-5}, \sqrt5\right )$ and $B'=\left (\sqrt{r^2-5}, \sqrt5\right )$ .

The gradient of $OA'=-\dfrac{\sqrt{5}}{\sqrt{r^2-5}}$ , so the gradient of $A'C'=\dfrac{\sqrt{r^2-5}}{\sqrt 5}\because$ the product of perpendicular gradients $=-1$ .

The line $A'C'$ is described by: $\begin{aligned}y-\sqrt 5&=\frac{\sqrt{r^2-5}}{\sqrt 5}\left (x+\sqrt{r^2-5}\right )\\\textnormal{At }C', x=0\implies y&=\frac{r^2-5}{\sqrt 5}+\sqrt 5\\\therefore C'&=\left (0, \frac{r^2-5}{\sqrt 5}+\sqrt 5\right )\end{aligned}$

The area of a kite is half the product of its diagonals, where $\left | A'B'\right |=2\sqrt{r^2-5}$ and $\left | OC'\right |=\dfrac{r^2-5}{\sqrt 5}+\sqrt 5$ .

We then have that:

$\begin{aligned}\textnormal{Area OACB}&=2\sqrt{\frac{11}{5}}r^2\\\iff \left (\frac{r^2-5}{\sqrt 5}+\sqrt 5\right )\left ( 2\sqrt{r^2-5}\right )&=2\sqrt{\frac{11}{5}}r^2\\\iff \frac{2\left (r^2-5\right )^{\frac{3}{2}}}{\sqrt{5}}+2\left (5r^2-25\right )^{\frac{1}{2}}&=2\sqrt{\frac{11}{5}}r^2\\\iff \frac{2\left (r^2-5\right )^{\frac{3}{2}}}{\sqrt{5}}+\frac{2\left (25r^2-125\right )^{\frac{1}{2}}}{\sqrt 5}&=\frac{2\sqrt{11}}{\sqrt{5}}r^2\\\iff 2\left (r^2-5\right )^\frac{3}{2}+2\left (25r^2-125\right )^{\frac{1}{2}}&=2\sqrt{11}r^2\\\iff \left (2r^2-10\right )\sqrt{r^2-5}+10\sqrt{r^2-5}&=2\sqrt{11}r^2\\\iff 2r^2\sqrt{r^2-5}&=2\sqrt{11}r^2\end{aligned}$

So that $r=0$ or $\sqrt{r^2-5}=\sqrt {11}\implies r^2=16\iff r=\pm 4$ .

Since we don't get a circle if $r=-4$ or $r=0$ , the radius $r$ of the circle is $\color{#20A900}{\boxed{4}}$ .