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(C) For the parabola ${y^2 = 8x}$,we have ${4a = 8}$,which gives ${a = 2}$.The vertex of the parabola is ${O(0, 0)}$.The endpoints of the latus rectum

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${x^2 + y^2 - 10x = 0}$

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(C) For the parabola ${y^2 = 8x}$,we have ${4a = 8}$,which gives ${a = 2}$.The vertex of the parabola is ${O(0, 0)}$.The endpoints of the latus rectum are ${L(a, 2a)}$ and ${L'(a, -2a)}$,which are ${L(2, 4)}$ and ${L'(2, -4)}$.Let the equation of the circle be ${x^2 + y^2 + 2gx + 2fy + c = 0}$.Since the circle passes through ${O(0, 0)}$,we have ${c = 0}$.Substituting ${L(2, 4)}$ into the equation: ${2^2 + 4^2 + 2g(2) + 2f(4) = 0 \implies 4 + 16 + 4g + 8f = 0 \implies 4g + 8f = -20 \implies g + 2f = -5}$.Substituting ${L'(2, -4)}$ into the equation: ${2^2 + (-4)^2 + 2g(2) + 2f(-4) = 0 \implies 4 + 16 + 4g - 8f = 0 \implies 4g - 8f = -20 \implies g - 2f = -5}$.Adding the two equations: ${2g = -10 \implies g = -5}$.Substituting ${g = -5}$ into ${g + 2f = -5}$,we get ${-5 + 2f = -5 \implies f = 0}$.Thus,the equation of the circle is ${x^2 + y^2 - 10x = 0}$.

The equation of a circle passing through the vertex and the extremities of the latus rectum of the parabola ${y^2 = 8x}$ is

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