Multiple choice

If one of the diameters of the circle ${ x }^{ 2 }+{ y }^{ 2 }-2x-6y+6=0$ is a chord to the circle with centre (2, 1), then the radius of the circle is

  1. 3

  2. 2

  3. 3/2

  4. 1
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A Correct answer
Explanation

The circle equation x^2 + y^2 - 2x - 6y + 6 = 0 has center (1, 3) and radius sqrt(1^2 + 3^2 - 6) = 2. A diameter of this circle is a chord of another circle with center (2, 1). The distance from (2, 1) to the line containing the diameter (which passes through (1, 3)) is needed. However, the diameter itself is a chord of the second circle, so the radius of the second circle R = sqrt(d^2 + r^2), where r=2 and d is the distance from (2, 1) to the center (1, 3). d = sqrt((2-1)^2 + (1-3)^2) = sqrt(1+4) = sqrt(5). R = sqrt(5 + 4) = 3.

AI explanation

The first circle's equation is rewritten as (x - 1)^2 + (y - 3)^2 = 4, showing its center is (1, 3) and its radius is 2. This circle's diameter, which has a length of 4, acts as a chord for the second circle centered at (2, 1). Using the perpendicular distance formula between the centers, d = sqrt((2 - 1)^2 + (1 - 3)^2) = sqrt(5). Applying the chord length formula 2 * sqrt(R^2 - d^2) = 4, we square both sides to get 4 * (R^2 - 5) = 16, which simplifies to R^2 - 5 = 4. Solving for R gives a radius of 3.

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