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QJEEM21Mae2dd: JEE Main 2021 Mathematics - Class 11 JEEM21Mae2dd Solution: Let ABCD be a square of side of unit length. Let a circle C 1…

Let ABCD be a square of side of unit length. Let a circle C1 centered at A with unit radius is drawn. Another circle C2 which touches C1 and the lines AD and AB are tangent to it, is also drawn. Let a tangent line from the point C to the circle C2 meet the side AB at E. If the length of EB is α\alpha + 3{\sqrt 3 } β\beta , where α\alpha , β\beta are integers, then α\alpha + β\beta is equal to ____________.

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JEE Main 2021 (Online) 16th March Morning Shift Mathematics - Properties of Triangle Question 18 English Explanation

(i) 2r+r=1\sqrt 2 r + r = 1

r=12+1r = {1 \over {\sqrt 2 + 1}}

r=2−1r = \sqrt 2 - 1

(ii) CC2=22−2=2(2−1)C{C_2} = 2\sqrt 2 - 2 = 2\left( {\sqrt 2 - 1} \right)

From ΔCC2N=sin⁡ϕ=2−12(2−1)\Delta C{C_2}N = \sin \phi = {{\sqrt 2 - 1} \over {2\left( {\sqrt 2 - 1} \right)}}

ϕ=30∘\phi = 30^\circ

(iii) In Δ\Delta ACE apply sine law

AEsin⁡ϕ=ACsin⁡105∘{{AE} \over {\sin \phi }} = {{AC} \over {\sin 105^\circ }}

AE=12×23+1.22AE = {1 \over 2} \times {{\sqrt 2 } \over {\sqrt 3 + 1}}.2\sqrt 2

AE=23+1=3−1AE = {2 \over {\sqrt 3 + 1}} = \sqrt 3 - 1

∴\therefore EB=1−(3−1)EB = 1 - \left( {\sqrt 3 - 1} \right)

= 2−32 - \sqrt 3

∴\therefore α\alpha = 2, β\beta = −- 1 ⇒\Rightarrow α\alpha + β\beta = 1

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