Multiple choice

The equation of the locus of the mid points of the chord of the circle $4\mathrm{x}^{2}+4\mathrm{y}^{2}-12\mathrm{x}+4\mathrm{y}+1=0$ that subtends an angle of $\displaystyle \frac{2\pi}{3}$ at its centre is

  1. $\displaystyle \mathrm{x}^{2}+\mathrm{y}^{2}-3\mathrm{x}+\mathrm{y}+\frac{16}{31}=0$
  2. $x^{2}+y^{2}-3x+y-\displaystyle \frac{31}{16}=0$
  3. $x^{2}+y^{2}+3x+y+\displaystyle \frac{31}{16}=0$
  4. $x^{2}+y^{2}-3x+y+\displaystyle \frac{31}{16}=0$
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D Correct answer
Explanation

The circle equation is x^2 + y^2 - 3x + y + 1/4 = 0. Center is (1.5, -0.5), radius squared is 2.25 + 0.25 - 0.25 = 2.25, so r=1.5. For a chord subtending 120 degrees, the distance from the center d = r * cos(60) = 1.5 * 0.5 = 0.75. The locus of the midpoint is a circle with radius d = 0.75. The equation is (x-1.5)^2 + (y+0.5)^2 = 0.75^2 = 0.5625 = 9/16. Expanding gives x^2 + y^2 - 3x + y + 2.25 + 0.25 - 0.5625 = 0, which is x^2 + y^2 - 3x + y + 1.9375 = 0, or 31/16.

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