A variable circle passes through the fixed po | vedclass

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(A) Let the coordinates of the fixed point $A$ be $(p, q)$ and the coordinates of the other end of the diameter $B$ be $(x, y)$.Since $AB$ is the diam

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$(x-p)^{2}=4 q y$

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(A) Let the coordinates of the fixed point $A$ be $(p, q)$ and the coordinates of the other end of the diameter $B$ be $(x, y)$.Since $AB$ is the diameter, the center of the circle is the midpoint of $AB$, which is $(\frac{x+p}{2}, \frac{y+q}{2})$.The radius of the circle is the distance from the center to $A$, which is $\sqrt{(\frac{x+p}{2}-p)^{2} + (\frac{y+q}{2}-q)^{2}} = \sqrt{(\frac{x-p}{2})^{2} + (\frac{y-q}{2})^{2}}$.Since the circle touches the $X$-axis, the radius is equal to the absolute value of the $y$-coordinate of the center, so $r = |\frac{y+q}{2}|$.Squaring both sides, we get $(\frac{x-p}{2})^{2} + (\frac{y-q}{2})^{2} = (\frac{y+q}{2})^{2}$.Multiplying by $4$, we get $(x-p)^{2} + (y-q)^{2} = (y+q)^{2}$.$(x-p)^{2} + y^{2} - 2qy + q^{2} = y^{2} + 2qy + q^{2}$.$(x-p)^{2} = 4qy$.

$A$ variable circle passes through the fixed point $A(p, q)$ and touches the $X$-axis. The locus of the other end of the diameter through $A$ is

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