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JEE Advanced 2012Paper 1
Exam
JEE Advanced
Year
2012
Paper
Paper 1
Subject
Mathematics
Chapter
Circle

JEE Advanced 2012 · Paper 1 · Mathematics · Circle

The locus of the mid-point of the chord of contact of tangents...

The locus of the mid-point of the chord of contact of tangents drawn from points lying on the straight line 4x−4 x- 5y=205 y=20 to the circle x2+y2=9x^{2}+y^{2}=9 isThe locus of the mid-point of the chord of contact of tangents drawn from points lying on the straight line 4 x- 5 y=20

Options

  1. A20(x2+y2)−36x+45y=020\left(x^{2}+y^{2}\right)-36 x+45 y=0
  2. B20(x2+y2)+36x−45y=020\left(x^{2}+y^{2}\right)+36 x-45 y=0
  3. C36(x2+y2)−20x+45y=036\left(x^{2}+y^{2}\right)-20 x+45 y=0
  4. D36(x2+y2)+20x−45y=036\left(x^{2}+y^{2}\right)+20 x-45 y=0

Correct answer

A. 20(x2+y2)−36x+45y=020\left(x^{2}+y^{2}\right)-36 x+45 y=0

Step-by-step solution

Any point PP on line 4x−5y=204 x-5 y=20 can be considered as \left(\alpha, \frac{4 \alpha-20}{5}\right).Equationofchordofcontact. Equation of chord of contactA Btothecircleto the circlex^{2}+y^{2}=9drawnfrompointdrawn from point\mathrm{P}\left(\alpha, \frac{4 \alpha-20}{5}\right)isx.is x.\alpha+y.y.\left(\frac{4 \alpha-20}{5}\right)=9...(i)<imgsrc="https://cdn−question−pool.getmarks.app/pyq/jeeadvanced/zkFIWBdkdHID8Ep1HUGZDpd4kkJTYXASn79Szm3a1E.original.fullsize.png"><br>Alsotheequationofchord<img src="https://cdn-question-pool.Quantrex.app/pyq/jee_advanced/zkFIWBdkdHID8Ep1HUGZDpd4kkJTYXASn79_Szm3a1E.original.fullsize.png"><br> Also the equation of chordA Bwhosemidpointiswhose mid point is(h, k)isish x+k y=h^{2}+k^{2}....(ii) \because \quadEquations(i)and(ii)representthesameline,Equations (i) and (ii) represent the same line,\therefore \quad \frac{h}{\alpha}=\frac{k}{\frac{4 \alpha-20}{5}}=\frac{h^{2}+k^{2}}{9} \Rightarrow 5 k \alpha=4 h \alpha-20 handand9 h=\alpha\left(h^{2}+k^{2}\right)⇒α=20h4h−5k\Rightarrow \alpha=\frac{20 h}{4 h-5 k} \quad and α=9hh2+k2\alpha=\frac{9 h}{h^{2}+k^{2}} ⇒20h4h−5k=9hh2+k2⇒20(h2+k2)=9(4h−5k)\Rightarrow \quad \frac{20 h}{4 h-5 k}=\frac{9 h}{h^{2}+k^{2}} \Rightarrow 20\left(h^{2}+k^{2}\right)=9(4 h-5 k) ∴\therefore \quad Locus of (h,k)(h, k) is 20(x2+y2)−36x+45y=020\left(x^{2}+y^{2}\right)-36 x+45 y=0
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