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JEE Main 2005
Exam
JEE Main
Year
2005
Subject
Mathematics
Chapter
Circle

JEE Main 2005 · Mathematics · Circle

If a circle passes through the point (a, b) and cuts the circle...

If a circle passes through the point (a,b)(a, b) and cuts the circle x2+y2=p2x^2+y^2=p^2 orthogonally, then the equation of the locus of its centre is

Options

  1. Ax2+y2−3ax−4by+(a2+b2−p2)=0x^2+y^2-3 a x-4 b y+\left(a^2+b^2-p^2\right)=0
  2. B2ax+2by−(a2−b2+p2)=02 a x+2 b y-\left(a^2-b^2+p^2\right)=0
  3. Cx2+y2−2ax−3by+(a2−b2−p2)=0x^2+y^2-2 a x-3 b y+\left(a^2-b^2-p^2\right)=0
  4. D2ax+2by−(a2+b2+p2)=02 a x+2 b y-\left(a^2+b^2+p^2\right)=0

Correct answer

D. 2ax+2by−(a2+b2+p2)=02 a x+2 b y-\left(a^2+b^2+p^2\right)=0

Step-by-step solution

Let the centre be (α,β)(\alpha, \beta) ∵\because It cut the circle x2+y2=p2\mathrm{x}^2+\mathrm{y}^2=\mathrm{p}^2 orthogonally 2(−α)×0+2(−β)×0=c1−p22(-\alpha) \times 0+2(-\beta) \times 0=c_1-p^2 c1=p2\mathrm{c}_1=\mathrm{p}^2 Let equation of circle is x2+y2−2αx−2βy+p2=0x^2+y^2-2 \alpha x-2 \beta y+p^2=0 It pass through (a,b)⇒a2+b2−2αa−2βb+p2=0(a, b) \Rightarrow a^2+b^2-2 \alpha a-2 \beta b+p^2=0 Locus $\therefore 2 a x+2 b y-\left(a^2+b^2+p^2\right)=0
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