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QJEEAdv19Madce0: JEE Advanced 2019 Mathematics - Class 11 JEEAdv19Madce0 Solution: Let the point B be the reflection of the point A(2, 3) with…

Let the point B be the reflection of the point A(2, 3) with respect to the line 8x−6y−23=08x - 6y - 23 = 0 . Let ΓA\Gamma_{A} and ΓB\Gamma_{B} be circles of radii 2 and 1 with centres A and B respectively. Let T be a common tangent to the circles ΓA\Gamma_{A} and ΓB\Gamma_{B} such that both the circles are on the same side of T. If C is the point of intersection of T and the line passing through A and B, then the length of the line segment AC is .................

Step-by-Step Verified Solution

According to given information the figure is as following

JEE Advanced 2019 Paper 1 Offline Mathematics - Circle Question 13 English Explanation

From the figure,

AC=2sin⁡θAC = {2 \over {\sin \theta }} ....(i)

∵\because sin⁡θ=1CB\sin \theta = {1 \over {CB}} (from Δ\Delta CPB) .... (ii)

and sin⁡θ=2AC=2CB+AB\sin \theta = {2 \over {AC}} = {2 \over {CB + AB}} (from Δ\Delta CQA) ....(iii)

∵\because AB = AM + MB = 2AM [ ∵\because AM = MB]

=2∣(8×2)−(6×3)−23∣64+36= 2{{\left| {(8 \times 2) - (6 \times 3) - 23} \right|} \over {\sqrt {64 + 36} }}

=2×2510=5.00= {{2 \times 25} \over {10}} = 5.00

From Eqs. (ii) and (iii), we get

sin⁡θ=1CB=2CB+AB\sin \theta = {1 \over {CB}} = {2 \over {CB + AB}}

⇒1CB=2CB+5\Rightarrow {1 \over {CB}} = {2 \over {CB + 5}} [ ∵\because AB = 5]

⇒CB+5=2CB⇒CB=5=1sin⁡θ\Rightarrow CB + 5 = 2CB \Rightarrow CB = 5 = {1 \over {\sin \theta }}

From the Eq. (i), we get

AC=2sin⁡θ=2×5=10.00AC = {2 \over {\sin \theta }} = 2 \times 5 = 10.00

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