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  • Let ABCD be a square of side of unit length. Let a circle centered at A with unit radius is drawn. Another circle which touches

Let ABCD be a square of side of unit length. Let a circle C_{1} centered at A with unit radius is drawn. Another circle C_{2} which touches C_{1}  and the lines AD and AB are tangent to it, is also drawn. Let a tangent line from the point C to the circle C_{2}  meet the side AB at E. If the length of EB is \alpha +\sqrt{3}\beta , where \alpha,\beta are integers, then \alpha+\beta is equal to _______________.
Option: 1 0
Option: 2 1
Option: 3 2
Option: 4 3

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best_answer

 

\begin{aligned} &\text {Here } AO + OD =1 \text { or }(\sqrt{2}+1) r =1 \\ &\Rightarrow \quad r =\sqrt{2}-1 \end{array}

\begin{aligned} &\text{Equation of circle }(x-r)^{2}+(y-r)^{2}=r^{2}\\ &\text{Equation of }CE\\ &y -1= m ( x -1)\\ &mx - y +1- m =0 \end{array}

\begin{aligned} &\text {It is tangent to circle }\\ \therefore&\left|\frac{ mr - r +1- m }{\sqrt{ m ^{2}+1}}\right|= r\\ &\left|\frac{( m -1) r +1- m }{\sqrt{ m ^{2}+1}}\right|= r\\ &\frac{(m-1)^{2}(r-1)^{2}}{m^{2}+1}=r^{2} \end{aligned}

\begin{aligned} &\text { Put } r =\sqrt{2}-1 \\ &\text { On solving } m =2-\sqrt{3}, 2+\sqrt{3} \end{array}

\begin{aligned} &\text { Taking greater slope of } CE \text { as }\\ &2+\sqrt{3}\\ &y-1=(2+\sqrt{3})(x-1)\\ &\text { Put } y =0\\ &-1=(2+\sqrt{3})( x -1)\\ &\frac{-1}{2+\sqrt{3}} \times\left(\frac{2-\sqrt{3}}{2-\sqrt{3}}\right)= x -1\\ &x -1=\sqrt{3}-1\\ &EB =1- x =1-(\sqrt{3}-1)\\ &EB =2-\sqrt{3} \end{aligned}

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Kuldeep Maurya

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