If y = 2x is a chord of the circle x2 + y2 = 10x, then equation of the circle whose diameter is the chord is
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x2 + y2 + 2x + 4y = 0
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x2 + y2 + 2x – 4y = 0
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x2 + y2 – 2x – 4y = 0
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None of these
The equation of a circle with chord y = 2x and circle x^2 + y^2 - 10x = 0 is given by S + kL = 0, where S is the circle and L is the line. Substituting the line into the circle equation gives the intersection points, but the standard formula for a circle with a given chord as diameter is S + lambda(L) = 0. Solving for the specific circle yields x^2 + y^2 - 2x - 4y = 0.
We substitute y = 2x into the original circle equation to find the intersection points, yielding x^2 + 4x^2 = 10x, which simplifies to x(x - 2) = 0, giving points (0, 0) and (2, 4). The equation of a circle with these points as the diameter is (x - x1)(x - x2) + (y - y1)(y - y2) = 0. Substituting the points gives (x - 0)(x - 2) + (y - 0)(y - 4) = 0, which expands to x^2 + y^2 - 2x - 4y = 0.