If two chords of the circle ${ x }^{ 2 }+{ y }^{ 2 }-ax-by=0$, drawn from the point $\left( a,b \right)$ is divided by the x-axis in the ratio $2:1$ then:
- ${ a }^{ 2 }>3{ b }^{ 2 }$
- ${ a }^{ 2 }v<3{ b }^{ 2 }$
- ${ a }^{ 2 }>{ 4b }^{ 2 }$
- ${ a }^{ 2 }<4{ b }^{ 2 }$
Parametrize the chord from (a,b) to its other intersection with the circle; for the chord to meet the x-axis in ratio 2:1, the point dividing it in ratio 2:1 must have y = 0. Enforcing the real intersection condition yields a^2 > 3b^2.
Let the point where the chord intersects the x-axis be (h, 0). Because this point divides the line segment from (a, b) to a point (x, y) on the circle in a 2:1 ratio, the section formula gives (a + x)/3 = h and (b + y)/3 = 0, meaning x = 3h - a. Because (x, y) lies on the circle x^2 + y^2 - ax - by = 0 and (a, b) also lies on this circle, substituting x = 3h - a and y = -b yields a quadratic equation in h of 9h^2 - 3ah = 0. Solving this provides the two roots h = 0 and h = a/3, meaning the x-axis must pass through the origin and (a/3, 0). Because (a/3, 0) lies inside the circle, its distance from the center (a/2, b/2) must be less than the radius sqrt((a^2 + b^2)/4), so taking the square of both sides of the inequality (a/6)^2 + (b/2)^2 < (a^2 + b^2)/4 reveals a^2 > 3b^2.