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JEE Advanced 2019Paper 1
Exam
JEE Advanced
Year
2019
Paper
Paper 1
Subject
Mathematics
Chapter
Straight Lines

JEE Advanced 2019 · Paper 1 · Mathematics · Straight Lines

Let the point B be the reflection of the point A(2, 3) with respect...

Let the point BB be the reflection of the point $A(2,3)A\left(2, 3\right) with respect to the line 8x−6y−23=0.8x - 6y - 23 = 0 . Let TAand{T}_{A} and{T}_{B}becircleofradiibe circle of radii2andand1withcentreswith centresAandandBrespectively.Letrespectively. LetTbeacommontangenttothecirclebe a common tangent to the circle{T}_{A}andand{T}_{B}suchthatboththecircleareonthesamesideofsuch that both the circle are on the same side ofT .IfIfCisthepointofintersectionofis the point of intersection ofTandthelinepassingthroughand the line passing throughAandandB,thenthelengthofthelinesegmentthen the length of the line segmentAC$ is ___________.Let the point B be the reflection of the point A(2, 3) with respect to the line 8x - 6y - 23 = 0 . Let T _ A and T _ BLet the point B be the reflection of the point A(2, 3) with respect to the line 8x - 6y - 23 = 0 . Let T _ A and T _ B

Correct answer

0

Step-by-step solution

Let the point B be the reflection of the point A(2, 3) with respect to the line 8x - 6y - 23 = 0 . Let T _ A and T _ B
Clearly,ACPACP and BCQBCQ are similar triangles.
Hence,
ACBC=r1r2\dfrac{AC}{BC} = \dfrac{{r}_{1}}{{r}_{2}}
ACBC=2\dfrac{AC}{BC} = 2
AC=2BCAC = 2BC ...(i)
Let the point B be the reflection of the point A(2, 3) with respect to the line 8x - 6y - 23 = 0 . Let T _ A and T _ B
Now, since BB , is image of AA in line 8x−6y=238x - 6y = 23
AD=BDAD = BD
AB=2ADAB = 2AD ...(ii)
AD=AD = distance of AA from line
8x−6y=238x - 6y = 23
AD=∣16−16−2310∣=52AD = \left|\dfrac{16 - 16 - 23}{10}\right| = \dfrac{5}{2}
Now as AC=2BCAC = 2BC
⇒AC=2(AC−AB)⇒ AC = 2 ( AC - AB )
⇒AC=2AB⇒ AC = 2AB
⇒AC=4AD⇒ AC = 4AD (from (ii))
AC=4×52AC = 4 \times \dfrac{5}{2}
AC=10AC = 10
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