Multiple choice

The locus of the point of intersection of the tangents at the ends of a chord of a circle $\mathrm{x}^{2}+ \mathrm{y}^{2}=\mathrm{a}^{2}$ which touches the circle $\mathrm{x}^{2}+\mathrm{y}^{2}-2\mathrm{a}\mathrm{x}=0$

  1. $\mathrm{y}^{2}=\mathrm{a} (a-2x)$
  2. $\mathrm{x}^{2}=\mathrm{a}(\mathrm{a}-2\mathrm{y})$
  3. $\mathrm{x}^{2}-\mathrm{y}^{2}=(\mathrm{x}-\mathrm{a})^{2}$
  4. $\mathrm{x}^{2}+\mathrm{y}^{2}=(\mathrm{y}-\mathrm{a})^{2}$
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A Correct answer
Explanation

The chord of x^2 + y^2 = a^2 touching x^2 + y^2 - 2ax = 0 is a standard geometry problem. The locus of the intersection of tangents at the ends of a chord is the polar of the intersection point. The result is y^2 = a(a-2x).

AI explanation

Let the point of intersection of the tangents be (h, k). The chord of contact from (h, k) to the circle x squared plus y squared equals a squared is hx + ky = a squared. Since this chord touches the circle x squared plus y squared minus 2ax equals 0, which has center (a, 0) and radius a, the perpendicular distance from the center to the line equals the radius. Setting up the equation absolute value of (ha plus 0k minus a squared) divided by the square root of (h squared plus k squared) equals a, we square and simplify to get h squared plus k squared equals 2ah. Replacing h and k with x and y yields y squared equals a multiplied by (a minus 2x).

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