Question

A circle \( S \equiv x^2 + y^2 + 2gx + 2fy + 6 = 0 \) cuts another circle \[ x^2 + y^2 - 6x - 6y - 6 = 0 \] orthogonally. If the angle between the circles \( S = 0 \) and \[ x^2 + y^2 + 6x + 6y + 2 = 0 \] is 60°, then the radius of the circle \( S = 0 \) is:

• A. \(2\)
• B. \(1\)
• C. \(4\)
• D. \(5\)

Hint:

Use the orthogonality condition and compare with the general circle equation to solve for unknowns.

Answer 1

The Correct Option is A

Let the circles be:
\(S_1: x^2 + y^2 + 2gx + 2fy + 6 = 0\)
\(S_2: x^2 + y^2 - 6x - 6y - 6 = 0\)
\(S_3: x^2 + y^2 + 6x + 6y + 2 = 0\)
\(S_1\) and \(S_2\) are orthogonal, so \(2g_1g_2 + 2f_1f_2 = c_1 + c_2\).
\(2g(-3) + 2f(-3) = 6 - 6\)
\(-6g - 6f = 0\)
\(g + f = 0\)
\(f = -g\)
The angle between \(S_1\) and \(S_3\) is 60°.
The centers of \(S_1\) and \(S_3\) are \(C_1(-g, -f)\) and \(C_3(-3, -3)\).
The radii of \(S_1\) and \(S_3\) are \(r_1 = \sqrt{g^2 + f^2 - 6}\) and \(r_3 = \sqrt{3^2 + 3^2 - 2} = \sqrt{16} = 4\).
The distance between the centers is \(d = \sqrt{(-g+3)^2 + (-f+3)^2}\).
Since the angle between the circles is 60°, we have:
\[\cos 60^\circ = \frac{r_1^2 + r_3^2 - d^2}{2r_1r_3}\] \[\frac{1}{2} = \frac{g^2 + f^2 - 6 + 16 - (g^2 - 6g + 9 + f^2 - 6f + 9)}{2r_1(4)}\] \[\frac{1}{2} = \frac{10 - (-6g - 6f + 18)}{8r_1}\] \[\frac{1}{2} = \frac{-8 + 6(g+f)}{8r_1}\] Since \(f = -g\), \(g + f = 0\). \[\frac{1}{2} = \frac{-8}{8r_1}\] \[\frac{1}{2} = \frac{-1}{r_1}\] \[r_1 = -2\] However, radius cannot be negative.
Let's use the formula with absolute value.
\[\cos 60^\circ = \frac{|r_1^2 + r_3^2 - d^2|}{2r_1r_3}\] \[\frac{1}{2} = \frac{|g^2 + f^2 - 6 + 16 - (g^2 - 6g + 9 + f^2 - 6f + 9)|}{8r_1}\] \[\frac{1}{2} = \frac{|-8 + 6(g+f)|}{8r_1}\] Since \(g+f = 0\),
\[\frac{1}{2} = \frac{|-8|}{8r_1}\] \[\frac{1}{2} = \frac{8}{8r_1} = \frac{1}{r_1}\] \[r_1 = 2\] Thus, the radius of \(S_1\) is 2. Final Answer: The final answer is $\boxed{(1)}$