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JEE Main 201219 MayOnline
Exam
JEE Main
Year
2012
Date
19 May
Mode
Online
Subject
Mathematics
Chapter
Properties of Triangles

JEE Main 2012 · 19 May · Online · Mathematics · Properties of Triangles

If three distinct points A, B, C are given in the 2dimensional...

If three distinct points A,B,CA, B, C are given in the 2dimensional coordinate plane such that the ratio of the distance of each one of them from the point (1,0)(1,0) to the distance from (−1,0)(-1,0) is equal to 12\frac{1}{2}, then the circumcentre of the triangle ABCA B C is at the point

Options

  1. A(53,0)\left(\frac{5}{3}, 0\right)
  2. B(0,0)(0,0)
  3. C(13,0)\left(\frac{1}{3}, 0\right)
  4. D(3,0)(3,0)

Correct answer

A. (53,0)\left(\frac{5}{3}, 0\right)

Step-by-step solution

Let P(1,0)andQ(−1,0),A(x,y)and Q(-1,0), A(x, y) Given: APAQ=BPBQ=CPCQ=12⇒2AP=AQ⇒4(AP)2=AQ2\begin{aligned} & \text{ Given: } \frac{A P}{A Q}=\frac{B P}{B Q}=\frac{C P}{C Q}=\frac{1}{2} \\ & \Rightarrow 2 A P=A Q \\ & \Rightarrow 4(A P)^2=A Q^2 \end{aligned}⇒4[(x−1)2+y2]=(x+1)2+y2⇒4(x2+1−2x)+4y2=x2+1+2x+y2⇒3x2+3y2−8x−2x+4−1=0⇒3x2+3y2−10x+3=0⇒x2+y2−103x+1=0\begin{aligned} & \Rightarrow 4\left[(x-1)^2+y^2\right]=(x+1)^2+y^2 \\ & \Rightarrow 4\left(x^2+1-2 x\right)+4 y^2=x^2+1+2 x+y^2 \\ & \Rightarrow 3 x^2+3 y^2-8 x-2 x+4-1=0 \\ & \Rightarrow 3 x^2+3 y^2-10 x+3=0 \\ & \Rightarrow x^2+y^2-\frac{10}{3} x+1=0 \end{aligned}$∴\therefore A lies on the circle given by (1). As BandCB and C also follow the same condition. ∴\therefore \quad Centre of circumcircle of △ABC=centreof\triangle A B C=centre ofcircle given by (1) =\left(\frac{5}{3}, 0\right) \text{. }$
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