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

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(D) Let $(h, k)$ be the center of the circle.Since the circle touches the $y$-axis at $(0, 2)$,the $y$-coordinate of the center must be $k = 2$,and th

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$(-4, 0)$

Vedclass · Answer

(D) Let $(h, k)$ be the center of the circle.Since the circle touches the $y$-axis at $(0, 2)$,the $y$-coordinate of the center must be $k = 2$,and the radius $r$ is $|h|$.The equation of the circle is $(x - h)^2 + (y - 2)^2 = h^2$.Since the circle passes through $(-1, 0)$,we substitute these coordinates into the equation:$(-1 - h)^2 + (0 - 2)^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 equation of the circle is $(x + \frac{5}{2})^2 + (y - 2)^2 = (\frac{5}{2})^2$.Expanding this,we get $x^2 + 5x + \frac{25}{4} + y^2 - 4y + 4 = \frac{25}{4}$,which simplifies to $x^2 + y^2 + 5x - 4y + 4 = 0$.Checking the options,for point $(-4, 0)$:$(-4)^2 + (0)^2 + 5(-4) - 4(0) + 4 = 16 + 0 - 20 - 0 + 4 = 0$.Since the point satisfies the equation,the circle passes through $(-4, 0)$.

The circle passing through the point $(-1, 0)$ and touching the $y$-axis at $(0, 2)$ also passes through the point

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