Question

If one of the diameters of the circle \(x^2 + y^2 - 2x - 6y + 6 = 0\) is a chord of another circle 'C', whose center is at (2, 1), then its radius is ________

Hint:

Use Pythagoras theorem: $R^2 = d^2 + r^2$, where $R$ is the radius of the outer circle, $r$ is the radius of the inner circle, and $d$ is the distance between their centers.

Answer 1

Correct Answer: 3

Step 1: Center of \(C_1\) is \(O_1(1, 3)\), radius \(r_1 = \sqrt{1+9-6} = 2\).
Step 2: The diameter of \(C_1\) (length 4) is a chord of circle \(C\). The midpoint of this chord is \(O_1(1, 3)\).
Step 3: Center of circle \(C\) is \(O(2, 1)\).
Step 4: Distance between centers \(d = \sqrt{(2-1)^2 + (1-3)^2} = \sqrt{1+4} = \sqrt{5}\).
Step 5: In \(\Delta OO_1P\) (where \(P\) is an endpoint of the chord): \(R^2 = d^2 + r_1^2 = (\sqrt{5})^2 + 2^2 = 5+4 = 9 \Rightarrow R = 3\).