Multiple choice

Consider a family of circles which are passing through the point $\left ( -1,1 \right )$ and are tangent to $x$-axis. If $\left ( h,k \right )$ are the coordinate of the centre of the circles, then the set of values of $k$ is given by the interval

  1. $\displaystyle -\frac{1}{2}\leq k\leq \displaystyle \frac{1}{2}$
  2. $k\leq \displaystyle \frac{1}{2}$
  3. $0\leq k\leq \displaystyle \frac{1}{2}$
  4. $k\geq \displaystyle \frac{1}{2}$
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D Correct answer
Explanation

A circle passing through (-1, 1) and tangent to the x-axis has a radius r = |k|. The equation is (x-h)^2 + (y-k)^2 = k^2. Substituting (-1, 1): (-1-h)^2 + (1-k)^2 = k^2. Expanding: 1 + 2h + h^2 + 1 - 2k + k^2 = k^2, so h^2 + 2h + 2 = 2k. Since h^2 + 2h + 2 = (h+1)^2 + 1, the minimum value is 1. Thus 2k >= 1, or k >= 1/2.

AI explanation

For a circle passing through (-1, 1) and tangent to the x-axis, its centre (h, k) must have a radius equal to its distance from the x-axis, meaning r = |k|. Using the distance formula between the centre and the point gives sqrt((h + 1)^2 + (k - 1)^2) = |k|. Squaring both sides yields (h + 1)^2 + k^2 - 2k + 1 = k^2, which simplifies to (h + 1)^2 = 2k - 1. Since the left side is a perfect square and must be non-negative, 2k - 1 >= 0, so k >= 1/2.

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