Question

Let $ABCD$ be a square of the side of unit length. Let a circle $C_1$ centered at $A$ with a unit radius is drawn. Another circle $C_2$ which touches $C_1$ and the lines $AD$ and $AB$ is tangent to it, is also drawn. Let a tangent line from point $C$ to the circle $C_2$ meet the side$AB$ at $E$. If the length of $EB$ is $ɑ+\sqrt{3}\beta$, where $ɑ,\beta$ are integers, then $ɑ+\beta$ is equal to ___

Answer 1

Explanation of the correct answer:

Illustrating the figure according to the given data :

Step:1 Finding radous of circle

From figure,

$AO+OD=1$

$$\begin{gathered} OA=\sqrt{2}r \\ OD=r \end{gathered}$$

since, we have $\therefore (\sqrt{2}+1)r=1$

$$\begin{gathered} \Rightarrow r=\frac{1}{(\sqrt{2}+1)} \\ \Rightarrow r=\frac{1}{(\sqrt{2}+1)}\left[\frac{(\sqrt{2}-1)}{(\sqrt{2}-1)}\right] \\ \Rightarrow r=(\sqrt{2}-1) \end{gathered}$$

So, the equation of circle be ${(x–r)}^2+{(y–r)}^2=r^2...\left(i\right)$

Step:2 Finding slope $m$ by equation of tangent

Equation of tangent $CE$ to the circle

$$\begin{gathered} \Rightarrow y–1=m(x–1)....\left(ii\right) \\ \Rightarrow mx–y+1–m=0...\left(iii\right) \end{gathered}$$

The perpendicular distance from Centre $(r,r)$ to the tangent line $\left(iii\right)$

$$\begin{gathered} \Rightarrow \left|\frac{mr-r+1-m}{\sqrt{m^2+1}}\right|=r \\ \Rightarrow \left|\frac{\left(m-1\right)r+1-m}{\sqrt{m^2+1}}\right|=r \\ \Rightarrow \frac{{\left(m-1\right)}^2{\left(r-1\right)}^2}{m^2+1}=r^2...\left(iv\right) \end{gathered}$$

Since, $r=\sqrt{2}-1$

Solving $\left(iv\right)$ by putting $r$ we have,

$m=2-\sqrt{3},2+\sqrt{3}$

Proceeding with greater value of slope

Then, $m=2+\sqrt{3}$

Step:3 Finding equation of tangent

Putting value $m$into $\left(ii\right)$

$$\begin{gathered} \Rightarrow y–1=\left(2+\sqrt{3}\right)(x–1) \\ \Rightarrow –1=\left(2+\sqrt{3}\right)(x–1) \\ \Rightarrow \frac{-1}{2+\sqrt{3}}\left[\frac{2-\sqrt{3}}{2-\sqrt{3}}\right]=x-1 \\ \Rightarrow x-1=\sqrt{3}-1 \end{gathered}$$

Therefore, From figure

$$\begin{gathered} EB=1-x=1-\left(\sqrt{3}-1\right) \\ \Rightarrow EB=2-\sqrt{3} \end{gathered}$$

Comparing it with the given expression $ɑ+\sqrt{3}\beta$ we have,

$ɑ=2,\beta =–1$

Hence, $ɑ+\beta =1$ is the answer.