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BITSAT 2022
Exam
BITSAT
Year
2022
Subject
Mathematics
Chapter
Circle

BITSAT 2022 · Mathematics · Circle

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

The locus of the mid-point of the chord if contact of tangents drawn from points lying on the straight line 4x−5y=204 x-5 y=20 to the circle x2+y2=9x^2+y^2=9 is

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

Let P(t,4t−205)beapointonP \left(t, \frac{4 t-20}{5}\right) be a point on the line 4 x-5 y=20. Then the chord of contact of tangents drawn from PP to the circle x2+y2=9x^2+y²=9 is tx+(4t−205)t x+\left(\frac{4 t-20}{5}\right) y=9\quad \quad \dots(i)Letθ(Let \theta( h, k)bethemid−pointofthischordofcontact,thenitsequationisalsobe the mid-point of this chord of contact, then its equation is alsoh x+k y=h²+k²\quad \quad \dots(ii) \left[\right.usingT=S′]using \left.T=S^{\prime}\right] Clearly Eqs. (i) and (ii) represent the same line ∴th=4t−205k=9h2+k2⇒th=9h2+k2 and th=4t−205k⇒t=9hh2+k2 and t=20h4h−5k⇒9hh2+k2=20h4h−5k⇒h{20(h2+k2)−36h+45k}=0x=0, or [20(x2+y2)−36x+45y]=0\begin{aligned} & \therefore \frac{t}{h}=\frac{4 t-20}{5 k}=\frac{9}{h^2+k^2} \\ & \Rightarrow \frac{t}{h}=\frac{9}{h^2+k^2} \text{ and } \frac{t}{h}=\frac{4 t-20}{5 k} \\ & \Rightarrow t=\frac{9 h}{h^2+k^2} \text{ and } t=\frac{20 h}{4 h-5 k} \\ & \Rightarrow \frac{9 h}{h^2+k^2}=\frac{20 h}{4 h-5 k} \\ & \Rightarrow h\left\{20\left(h^2+k^2\right)-36 h+45 k\right\}=0 \\ & x=0, \text{ or }\left[20\left(x^2+y^2\right)-36 x+45 y\right]=0 \end{aligned}
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