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(B) The equation of a circle with center $(h, k)$ tangent to the $x$-axis is $(x-h)^{2} + (y-k)^{2} = k^{2}$.Since the circle passes through the point

Vedclass · Accepted Answer

$k \ge \frac{1}{2}$

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(B) The equation of a circle with center $(h, k)$ tangent to the $x$-axis is $(x-h)^{2} + (y-k)^{2} = k^{2}$.Since the circle passes through the point $(-1, 1)$,we substitute these coordinates into the equation:$(-1-h)^{2} + (1-k)^{2} = k^{2}$$1 + 2h + h^{2} + 1 - 2k + k^{2} = k^{2}$$h^{2} + 2h + 2 - 2k = 0$For $h$ to be a real coordinate,the discriminant $D$ of this quadratic equation in $h$ must be greater than or equal to $0$:$D = (2)^{2} - 4(1)(2 - 2k) \ge 0$$4 - 8 + 8k \ge 0$$8k - 4 \ge 0$$8k \ge 4$$k \ge \frac{1}{2}$

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

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