If a circle passing through the point (-1, 0) | vedclass

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(B) Let the center of the circle be $(h, 2)$ and its radius be $|h|$. Since the circle touches the $y-$ axis at $(0, 2)$,the distance from the center

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$3$

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(B) Let the center of the circle be $(h, 2)$ and its radius be $|h|$. Since the circle touches the $y-$ axis at $(0, 2)$,the distance from the center $(h, 2)$ to the $y-$ axis is $|h|$.Since the circle passes through $(-1, 0)$,the distance from the center $(h, 2)$ to $(-1, 0)$ must equal the radius $|h|$.Thus,$(h - (-1))^2 + (2 - 0)^2 = h^2$$(h + 1)^2 + 4 = h^2$$h^2 + 2h + 1 + 4 = h^2$$2h + 5 = 0 \Rightarrow h = -\frac{5}{2}$.The center is $(-\frac{5}{2}, 2)$ and the radius is $r = |h| = \frac{5}{2}$.The equation of the circle is $(x + \frac{5}{2})^2 + (y - 2)^2 = (\frac{5}{2})^2$.To find the chord along the $x-$ axis,set $y = 0$:$(x + \frac{5}{2})^2 + (0 - 2)^2 = \frac{25}{4}$$(x + \frac{5}{2})^2 + 4 = \frac{25}{4}$$(x + \frac{5}{2})^2 = \frac{25}{4} - 4 = \frac{9}{4}$$x + \frac{5}{2} = \pm \frac{3}{2}$.So,$x_1 = -\frac{5}{2} + \frac{3}{2} = -1$ and $x_2 = -\frac{5}{2} - \frac{3}{2} = -4$.The length of the chord is $|x_1 - x_2| = |-1 - (-4)| = 3$.

If a circle passing through the point $(-1, 0)$ touches the $y-$ axis at $(0, 2)$,then the length of the chord of the circle along the $x-$ axis is

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