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JEE Main 2009
Exam
JEE Main
Year
2009
Subject
Mathematics
Chapter
Straight Lines

JEE Main 2009 · Mathematics · Straight Lines

Three distinct points A, B and C are given in the 2 - dimensional...

Three distinct points A, B and C are given in the 2 - dimensional coordinate plane such that the ratio of the distance of any one of them from the point (1,0)(1,0) to the distance from the point (−1,0)(-1,0) is equal to 13\frac{1}{3}. Then the circumcentre of the triangle ABCA B C is at the point

Options

  1. A(0,0)(0,0)
  2. B(54,0)\left(\frac{5}{4}, 0\right)
  3. C(52,0)\left(\frac{5}{2}, 0\right)
  4. D(53,0)\left(\frac{5}{3}, 0\right)

Correct answer

B. (54,0)\left(\frac{5}{4}, 0\right)

Step-by-step solution

P=(1,0);Q(−1,0) Let A=(x,y)APAQ=BPBQ=CPCQ=13⇒3AP=AQ⇒9AP2=AQ2⇒9(x−1)2+9y2=(x+1)2+y2⇒9x2−18x+9+9y2=x2+2x+1+y2⇒8x2−20x+8y2+8=0⇒x2+y2−52x+1=0\begin{aligned} & P=(1,0) ; Q(-1,0) \\ & \text{ Let } A=(x, y) \\ & \frac{A P}{A Q}=\frac{B P}{B Q}=\frac{C P}{C Q}=\frac{1}{3} \\ & \Rightarrow 3 A P=A Q \Rightarrow 9 A P^2=A Q^2 \Rightarrow 9(x-1)^2+9 y^2=(x+1)^2+y^2 \\ & \Rightarrow 9 x^2-18 x+9+9 y^2=x^2+2 x+1+y^2 \Rightarrow 8 x^2-20 x+8 y^2+8=0 \\ & \Rightarrow x^2+y^2-\frac{5}{2} x+1=0 \end{aligned} ∴\therefore A lies on the circle Similarly B, C are also lies on the same circle ∴\therefore Circumcentre of ABC=\mathrm{ABC}= Centre of Circle (1)=\left(\frac{5}{4}, 0\right)
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