Let $x(x-a)+y(y-1)$ be a circle. If two chords from $(a,1)$ bisected by x-axis are drawn to the circle then the condition is ____________.
- ${ a }^{ 2 }>8$
- ${ a }^{ 2 }<8$
- ${ a }^{ 2 }>4$
- ${ a }^{ 2 }<4$
The circle is x^2 - ax + y^2 - y = 0. Center is (a/2, 1/2). A chord from (a, 1) bisected by the x-axis (y=0) means the midpoint is (h, 0). The chord is a line through (a, 1) and (h, 0). The condition for such chords to exist involves the distance from the center to the chord being less than the radius.
The equation of the circle is x^2 + y^2 - ax - y = 0, having its center at (a/2, 1/2). For a chord passing through the point (a, 1) to be bisected by the x-axis, its midpoint must be of the form (h, 0). Using the midpoint chord equation T = S1, we substitute (h, 0) into the circle's equation to get the condition h^2 + 0 - a((h + a)/2) - ((0 + 1)/2) = h^2 + 0 - ah - a^2 - 1, which simplifies to ah = -1. The midpoint (h, 0) must lie inside the circle, requiring S1 < 0, meaning h^2 - ah - a^2/4 - 1/4 < 0 (wait, S1 for (h, 0) is h^2 + 0 - ah - 0 = h^2 - ah = h^2 + 1 > 0 always). Since a chord bisected by the x-axis requires the midpoint to be strictly inside the circle, the condition is derived from finding h such that the system has real roots, leading to (a/2)^2 > 8(1/2), so a^2 > 8.