The equation of the circle with the chord $y=2x$ of the circle $x^{2}+y^{2}-10x=0$ as its diameter is
- $x^{2}+y^{2}-2x-4y-5=0$
- $x^{2}+y^{2}=2x+4y$
- $x^{2}+y^{2}=4x+2y$
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none of these
The circle is x^2 + y^2 - 10x = 0. The intersection of this circle and the line y = 2x gives the endpoints of the diameter. Substituting y = 2x into the circle equation: x^2 + (2x)^2 - 10x = 0, so 5x^2 - 10x = 0, meaning x = 0 or x = 2. The points are (0,0) and (2,4). The equation of a circle with diameter endpoints (x1, y1) and (x2, y2) is (x-x1)(x-x2) + (y-y1)(y-y2) = 0. This gives (x-0)(x-2) + (y-0)(y-4) = 0, which is x^2 - 2x + y^2 - 4y = 0, or x^2 + y^2 = 2x + 4y.
Solving the system of equations x squared plus y squared minus 10x equals 0 and y equals 2x gives the endpoints of the diameter as (0, 0) and (2, 4). The equation of a circle with (x1, y1) and (x2, y2) as endpoints of a diameter is (x minus x1) times (x minus x2) plus (y minus y1) times (y minus y2) equals 0. Substituting the points gives x times (x minus 2) plus y times (y minus 4) equals 0, which simplifies to x squared plus y squared minus 2x minus 4y equals 0, or x squared plus y squared equals 2x plus 4y.