Question

If a diameter of the circle $x^2 + y^2 - 2x - 6y + 6 = 0$ is a chord of another circle $C$ having centre $(2, 1)$, then the radius of the circle $C$ is:

• A. $2$
• B. $\sqrt{3}$
• C. $3$
• D. $\sqrt{5}$
• E. $5$

Hint:

Visualize the geometry: the distance from the center of circle $C$ to the midpoint of its chord (which is the center of $S_1$) is the "perpendicular distance." The radius of $S_1$ is half the chord length.

Answer 1

The Correct Option is C

Concept: When a diameter of circle $S_1$ is a chord of circle $C$, the endpoints of that diameter lie on circle $C$. The center of $S_1$ is the midpoint of this chord. We can use the Pythagorean theorem on the triangle formed by the center of $C$, the center of $S_1$, and an endpoint of the chord.

Step 1: Find the center and radius of the first circle.
Equation: $x^2 + y^2 - 2x - 6y + 6 = 0$.
Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$:
• Center $C_1 = (-g, -f) = (1, 3)$.

• Radius $r_1 = \sqrt{g^2 + f^2 - c} = \sqrt{(-1)^2 + (-3)^2 - 6} = \sqrt{1 + 9 - 6} = 2$.

Step 2: Calculate the distance between the two centers.
Center of Circle $C$ is $C_2(2, 1)$. Center of $S_1$ is $C_1(1, 3)$. \[ d = \sqrt{(2 - 1)^2 + (1 - 3)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{5} \]

Step 3: Apply the Pythagorean theorem to find radius $R$ of Circle $C$.
The center $C_2$, center $C_1$, and an endpoint of the diameter form a right triangle where $R$ is the hypotenuse: \[ R^2 = d^2 + r_1^2 \] \[ R^2 = (\sqrt{5})^2 + 2^2 = 5 + 4 = 9 \] \[ R = 3 \]