Multiple choice

The locus of the midpoints of the chords of the circle $\left ({ x }^{ 2 }+{ y }^{ 2 }-ax-by\right )=0$ which subtend a right angle at $\left (\dfrac {a}{2},\dfrac {b}{2}\right )$ is

  1. $ax+by=0$
  2. $ax+by={ a }^{ 2 }+{ b }^{ 2 }$
  3. ${ x }^{ 2 }+{ y }^{ 2 }-ax-by+\cfrac { { a }^{ 2 }+{ b }^{ 2 } }{ 8 } =0$
  4. ${ x }^{ 2 }+{ y }^{ 2 }-ax-by-\cfrac { { a }^{ 2 }-{ b }^{ 2 } }{ 8 } =0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The locus of midpoints of chords subtending a right angle at a point is derived from the circle geometry properties. The correct equation is x^2 + y^2 - ax - by + (a^2 + b^2)/8 = 0.

AI explanation

Let the chord endpoints be X and Y, the midpoint be M, and the given point P be (a/2, b/2). The circle equation is x^2 + y^2 - ax - by = 0, and the angle XPY is 90 degrees. The radius R is given by R^2 = (a/2)^2 + (b/2)^2 = (a^2 + b^2)/4. Using the property of right triangles, PM = R / sqrt(2) = sqrt(a^2 + b^2) / (2*sqrt(2)). Squaring this distance gives PM^2 = (a^2 + b^2) / 8. The locus of midpoint M is given by the distance formula PM^2 = (x - a/2)^2 + (y - b/2)^2 = (a^2 + b^2)/8. Expanding this equation yields x^2 + y^2 - ax - by + (a^2 + b^2)/8 = 0. The correct option is C.

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