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Circle question

2009 · Shift 0 · Q39
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  5. /2009 · Shift 0 · Q39

Circle question

2009 · Shift 0 · Q39

JEE MainMathematicsCircleMCQ+4 / −1
Three distinct points A, B and C are given in the 2 -dimensional coordinates plane such that the ratio of the distance of any one of them from the point (1,0)(1, 0)(1,0) to the distance from the point (−1,0)(-1, 0)(−1,0) is equal to 13{1 \over 3}31​. Then the circumcentre of the triangle ABC is at the point :
  1. A
    (54,0)\left( {{5 \over 4},0} \right)(45​,0)
  2. B
    (52,0)\left( {{5 \over 2},0} \right)(25​,0)
  3. C
    (53,0)\left( {{5 \over 3},0} \right)(35​,0)
  4. D
    (0,0)\left( {0,0} \right)(0,0)
View written solutionFree

Correct answer: A

  1. Let any one of the points be P(x,y)P(x,y)P(x,y).
    Given, PA?\frac{PA}{?}?PA​ More precisely, the condition is that for each of the points A,B,CA,B,CA,B,C, distance from (1,0)distance from (−1,0)=13.\frac{\text{distance from }(1,0)}{\text{distance from }(-1,0)}=\frac13.distance from (−1,0)distance from (1,0)​=31​. So for a general point P(x,y)P(x,y)P(x,y) on the locus, (x−1)2+y2(x+1)2+y2=13.\frac{\sqrt{(x-1)^2+y^2}}{\sqrt{(x+1)^2+y^2}}=\frac13.(x+1)2+y2​(x−1)2+y2​​=31​.

  2. Squaring both sides, (x−1)2+y2(x+1)2+y2=19.\frac{(x-1)^2+y^2}{(x+1)^2+y^2}=\frac19.(x+1)2+y2(x−1)2+y2​=91​. Hence, 9((x−1)2+y2)=(x+1)2+y2.9\big((x-1)^2+y^2\big)=(x+1)^2+y^2.9((x−1)2+y2)=(x+1)2+y2.

  3. Expand: 9(x2−2x+1+y2)=x2+2x+1+y2.9(x^2-2x+1+y^2)=x^2+2x+1+y^2.9(x2−2x+1+y2)=x2+2x+1+y2. 9x2−18x+9+9y2=x2+2x+1+y2.9x^2-18x+9+9y^2=x^2+2x+1+y^2.9x2−18x+9+9y2=x2+2x+1+y2. 8x2−20x+8+8y2=0.8x^2-20x+8+8y^2=0.8x2−20x+8+8y2=0. Divide by 444: 2x2−5x+2+2y2=0.2x^2-5x+2+2y^2=0.2x2−5x+2+2y2=0.

  4. Rewrite as x2+y2−52x+1=0.x^2+y^2-\frac52 x+1=0.x2+y2−25​x+1=0. Complete the square in xxx: (x−54)2+y2=916.\left(x-\frac54\right)^2+y^2=\frac{9}{16}.(x−45​)2+y2=169​.

  5. Thus the locus of each of the points A,B,CA,B,CA,B,C is the circle with centre (54,0).\left(\frac54,0\right).(45​,0).

  6. Since A,B,CA,B,CA,B,C are three distinct points on this same circle, the circumcircle of triangle ABCABCABC is exactly this circle. Therefore, its circumcentre is (54,0).\left(\frac54,0\right).(45​,0).

  7. Checking options:

    • A: (54,0)\left(\frac54,0\right)(45​,0) ✓
    • B, C, D are incorrect.

Therefore, the correct answer is Option A.

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