Find the radius of the circle given in the picture below.

Question

This is the image of the question. I am not able to get how to find the radius. Please help with that.

This is my try. I can't proceed now after it.

Thanks

Answer 1

$$PT^2=PC\cdot PA\implies PT^2=2\cdot 5=10$$

Using Pythagorean theorem in right $\Delta PTO$, $PO=\sqrt{OT^2+PT^2}=\sqrt{R^2+10}$

Using cosine rule in $\Delta ABP$, $$\cos\angle APB=\frac{PB^2+PA^2-AB^2}{2(PB)(PA)}=\frac{(\sqrt{R^2+10}+R)^2+5^2-(\sqrt{11})^2}{2(\sqrt{R^2+10}+R)(5)}\tag1$$ Similarly, using cosine rule in $\Delta AOP$, $$\cos\angle APB=\frac{PO^2+PA^2-OA^2}{2(PO)(PA)}=\frac{(\sqrt{R^2+10})^2+5^2-(R)^2}{2(\sqrt{R^2+10})(5)}\tag2$$ Equating the values of $\cos \angle APB$ from (1) & (2), $$\frac{(\sqrt{R^2+10}+R)^2+5^2-(\sqrt{11})^2}{2(\sqrt{R^2+10}+R)(5)}=\frac{(\sqrt{R^2+10})^2+5^2-(R)^2}{2(\sqrt{R^2+10})(5)}$$ $$\frac{2R\sqrt{R^2+10}+2R^2+24}{\sqrt{R^2+10}+R}=\frac{35}{\sqrt{R^2+10}}$$ $$(11-2R^2)\sqrt{R^2+10}=2R^3-15R$$ $$(11-2R^2)^2(\sqrt{R^2+10})^2=(2R^3-15R)^2$$ $$28R^4-272R^2+605=0$$ $$R^2=\frac{-(-272)\pm\sqrt{(-272)^2-4\cdot 28\cdot605}}{2\cdot 56}=\frac{68\pm\sqrt{389}}{14}$$ $$R=\sqrt{\frac{68+\sqrt{389}}{14}}\approx 2.50318487$$

Answer 2

Given: $CD=2,\,BD=3,\,AB=\sqrt{11}$.
Let's define $\angle B=\angle DBE,\, CE=d$.
$2R=\frac{\sqrt{11}}{\cos A}$, $BE=\frac{\sqrt{11}}{\sin A}, \angle DBE=\angle DAE$, thus $CBE\sim CAD$ $$\frac{CB}{CA}=\frac{BE}{AD}=\frac{CE}{CD}$$ $$\frac{5}{d+2R}=\frac{\frac{\sqrt{11}}{\sin A}}{AD}=\frac{d}{2}$$ By applying the cosine rule to both $\Delta CBA$, $\Delta CDA$ we get $$\begin{cases} 11=25+(2R+d)^2-2\cdot 5\cdot (2R+d)\cos C,\\ AD^2=4+(2R+d)^2-2\cdot 2\cdot (2R+d)\cos C, \end{cases}$$ By eliminating $\cos C$ we get $$ AD^2=4+(2R+d)^2-\frac{2}{5}\left(25+(2R+d)^2-11\right)=$$ $$\frac{3}{5}(2R+d)^2-\frac{12}{5}=15\left(\frac{2R+d}{5}\right)^2-\frac{12}{5}=\frac{60}{d^2}-\frac{12}{5}$$ Thus we get $$\begin{cases} 10=d(d+2R)\\ 2\sqrt{11}=d\sqrt{\frac{60}{d^2}-\frac{60}{25}}\sin A \end{cases}$$ $$\begin{cases} 10=d(d+\frac{\sqrt{11}}{\cos A})\\ 2\sqrt{11}=\frac{\sqrt{60}}{5}\sqrt{25-d^2}\sin A \end{cases}$$ hence we get $d$ and then $R$.

Answer 3

\begin{align} x&=|OA|-|OE| =\sqrt{|AF|^2+|OF|^2}-|OE| \\ &=\sqrt{|AF|^2+|OC|^2-|CF|^2}-|OE| =\sqrt{R^2+10}-R \tag{1}\label{1} . \end{align}

By the Stewart's theorem

\begin{align} \triangle ABC:\quad |AC|^2\cdot|OB| + |BC|^2\cdot|OA| - |AB|\cdot(|OC|^2+|OA|\cdot|OB|) &=0 ,\\ 25\,R+11\,(x+R)-(x+2R)(R^2+R\,(x+R))&=0 ,\\ 25\,R+11\,\sqrt{R^2+10}-(\sqrt{R^2+10}+R)(R^2+R\,\sqrt{R^2+10})&=0 \tag{2}\label{2} ,\\ -\sqrt{R^2+10}(2R^2-11)+15R-2R^3&=0 ,\\ (R^2+10)(2R^2-11)^2-(15R-2R^3)^2&=0 ,\\ 28R^4-272R^2+605&=0 \tag{3}\label{3} , \end{align}
and the only suitable root of \eqref{3} is

\begin{align} R&=\tfrac1{14}\,\sqrt{952+14\sqrt{389}} \approx 2.503184870 . \end{align}

Answer 4

Following the notation of Alexey Burdin, we have two diametrically opposite points $A$ and $E$ on a circle of radius $R$, and two further points $B$ and $D$ on the circle such that $AE$ produced meets $BD$ produced at a point $C$ outside the circle. We are given that $|AB|=\surd11$, $|BD|=3$, and $|DC|=2$. We further write $d:=|EC|$ and $\theta:=\angle AEB$.

Now $\angle EBA$ is a right angle, whence $|EB|=\surd(4R^2-11)$ and $$\cos\theta=\frac{\surd(4R^2-11)}{2R}.$$Applying the cosine rule to $\triangle ECB$ gives $d^2+4R^2-11+2d\surd(4R^2-11)\cos\theta=(2+3)^2$ or, after substitution for $\cos\theta$, $$d^2+\frac{4R^2-11}Rd=36-4R^2.\qquad(1)$$Since $ABDE$ is cyclic, we have $|CE||CA|=|CD||CB|$; namely$$\qquad\qquad d(d+2R)=10.\qquad \qquad(2)$$After subtracting eqn $2$ from eqn $1$ to cancel $d^2$, we get$$d=\frac{2R(13-2R^2)}{2R^2-11}.$$ Now substituting this expression for $d$ back into eqn 2 yields$$\frac{8R^2(13-2R^2)}{(2R^2-11)^2}=10,$$which simplifies to$$28R^4-272R^2+605=0.$$The quadratic formula (for $R^2$) then gives$$R=\sqrt\frac{68+\surd389}{14}\approx2.50318487.$$(We reject the other root as it makes $d$ negative.)