Let the point B be the reflection of the point A(2, 3) with respect to the line . Let and be circles of radii 2 and 1 with centers A and B respectively. Let T be a common tangent to the circles and such that both the circles are on the same side of T.

Question

Let the point B be the reflection of the point A(2, 3) with respect to the line $8 x - 6 y - 23 = 0$ . Let ${I^{-}}_{A}$ and ${I^{-}}_{B}$ be circles of radii 2 and 1 with centers A and B respectively. Let T be a common tangent to the circles ${I^{-}}_{A}$ and ${I^{-}}_{B}$ 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 _______ . [2019]

Smart Achievers · Accepted Answer

(10)

$A N =$ $| \frac{16 - 18 - 23}{\sqrt{64 + 36}} |$ $= \frac{25}{10} = \frac{5}{2} = B N$

$∵$ $∆ C P A ~ ∆ C Q B (By AA similarity)$

$∴$ $\frac{C A}{C B} = \frac{P A}{Q B} \Rightarrow \frac{C A}{C A - 5} = \frac{2}{1}$

$\Rightarrow C A = 2 C A - 10 \Rightarrow C A = 10$

Solution

Ans.
(10)

$A N =$ $| \frac{16 - 18 - 23}{\sqrt{64 + 36}} |$ $= \frac{25}{10} = \frac{5}{2} = B N$

$∵$ $∆ C P A ~ ∆ C Q B (By AA similarity)$

$∴$ $\frac{C A}{C B} = \frac{P A}{Q B} \Rightarrow \frac{C A}{C A - 5} = \frac{2}{1}$

$\Rightarrow C A = 2 C A - 10 \Rightarrow C A = 10$

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Solution

Ans. (10) AN=|16-18-2364+36|=2510=52=BN ∵ ∆CPA~∆CQB (By AA similarity) ∴ CACB=PAQB ⇒ CACA-5=21 ⇒CA=2CA-10 ⇒ CA=10

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Ans.
(10)

$A N =$ $| \frac{16 - 18 - 23}{\sqrt{64 + 36}} |$ $= \frac{25}{10} = \frac{5}{2} = B N$

$∵$ $∆ C P A ~ ∆ C Q B (By AA similarity)$

$∴$ $\frac{C A}{C B} = \frac{P A}{Q B} \Rightarrow \frac{C A}{C A - 5} = \frac{2}{1}$

$\Rightarrow C A = 2 C A - 10 \Rightarrow C A = 10$