The tangent to the circle $C_{1} : x^{2} + y^{2} - 2x - 1 = 0$ at the point $(2, 1)$ cuts off a chord of length $4$ from a circle $C_{2}$ whose centre is $(3, -2)$. The radius of $C_{2}$ is

The center of circle C1, given by x^2 + y^2 - 2x - 1 = 0, is at (1, 0). The slope of the radius to the point (2, 1) is (1 - 0) / (2 - 1) = 1. The tangent line at (2, 1) is perpendicular to this radius, so its slope is -1. The equation of this tangent line is y - 1 = -1(x - 2), which simplifies to y = -x + 3, or x + y - 3 = 0. We calculate the perpendicular distance from the center of C2 (3, -2) to this tangent line using the point-to-line distance formula: |3 + (-2) - 3| / sqrt(1^2 + 1^2) = |-2| / sqrt(2) = 2/sqrt(2) = sqrt(2). Let r be the radius of circle C2. The line cuts a chord of length 4 from C2, so half the chord length is 2. Using the property of right triangles formed in a circle by the radius, half-chord, and perpendicular distance from the center, we have r^2 = (perpendicular distance)^2 + (half-chord length)^2. Substituting the known values gives r^2 = (sqrt(2))^2 + 2^2 = 2 + 4 = 6. Therefore, the radius r is sqrt(6). The result is sqrt(6).

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