Let the circles pass through the point (-1, 1 | vedclass

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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

$k \geq 1/2$

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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 have $(-1 - h)^2 + (1 - k)^2 = k^2$.Expanding this,we get $(h + 1)^2 + 1 - 2k + k^2 = k^2$,which simplifies to $h^2 + 2h + 2 - 2k = 0$.This is a quadratic equation in $h$: $h^2 + 2h + (2 - 2k) = 0$.For $h$ to be a real value,the discriminant $D$ must be greater than or equal to $0$.$D = b^2 - 4ac = (2)^2 - 4(1)(2 - 2k) \geq 0$.$4 - 8 + 8k \geq 0$.$8k - 4 \geq 0$.$8k \geq 4$.$k \geq 1/2$.

Let the circles pass through the point $(-1, 1)$ and be tangent to the $x$-axis. If $(h, k)$ are the coordinates of the center of the circle,then the set of values of $k$ is given by which interval?

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