The centre of the circle passing through $(6,0)$ and tangent to the circle $x^{2}+y^{2}=4$ at $(2, 0)$ is
- $(0,-6)$
- $(1,-5)$
- $(0,-9)$
-
none
The circle x^2 + y^2 = 4 has center (0,0) and radius 2. A circle tangent to it at (2,0) must have its center on the x-axis. Since it also passes through (6,0), the center must be the midpoint of the chord or satisfy the distance condition. Any circle tangent at (2,0) must have a center at (c, 0). The distance from (c,0) to (2,0) is |2-c| = radius. The distance from (c,0) to (6,0) is |6-c| = radius. Thus |2-c| = |6-c|, which has no solution (2-c = 6-c implies 2=6).
The new circle and the circle x squared plus y squared equals 4 are both tangent to the x-axis at the point (2, 0), meaning their centers must lie on the vertical line x equals 2. Because the new circle also passes through (6, 0), the distance from its center (2, k) to (2, 0) must equal its distance to (6, 0). Setting the absolute value of k equal to the square root of the sum of (2 minus 6) squared and (k minus 0) squared gives the equation k squared equals 16 plus k squared, which is impossible. Because no such center can exist, the correct choice is none of the provided options.