Question

A circle C touches the line \(x=2y\) at the point (2, 1) and intersects the circle C\(_1\) : \(x^2 + y^2 + 2y - 5 = 0\) at two points P and Q such that PQ is a diameter of C\(_1\). Then the diameter of C is :

• A. 15
• B. \(4\sqrt{15}\)
• C. \(\sqrt{285}\)
• D. \(7\sqrt{5}\)

Hint:

For problems involving intersecting circles and tangents, systematically list the geometric conditions and translate them into algebraic equations. The common chord is S-S'=0, the radius is perpendicular to the tangent at the point of contact, and so on. Careful algebraic manipulation is key.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We have a circle C that is tangent to a line at a given point. It also intersects another circle C\(_1\) such that their common chord is a diameter of C\(_1\). We need to find the diameter of circle C.
Step 2: Key Formula or Approach:
1. The common chord of two intersecting circles is given by the equation S - S' = 0, where S=0 and S'=0 are the equations of the two circles.
2. The center of circle C must lie on the line perpendicular to the tangent \(x=2y\) at the point of tangency (2, 1).
3. The common chord PQ passes through the center of C\(_1\).
Step 3: Detailed Explanation:
Analyze Circle C\(_1\):
The equation of C\(_1\) is \(x^2 + y^2 + 2y - 5 = 0\).
Center of C\(_1\), let's call it O\(_1\), is \((0, -1)\).
Radius of C\(_1\), r\(_1\), is \(\sqrt{g^2 + f^2 - c} = \sqrt{0^2 + 1^2 - (-5)} = \sqrt{6}\).
Since PQ is the diameter of C\(_1\), the line PQ is the common chord and it must pass through the center of C\(_1\), which is (0, -1).
Equation of Circle C:
Let the equation of circle C be \(x^2 + y^2 + 2gx + 2fy + c = 0\).
The common chord PQ has the equation S - S\(_1\) = 0:
\( (x^2 + y^2 + 2gx + 2fy + c) - (x^2 + y^2 + 2y - 5) = 0 \)
\( 2gx + (2f-2)y + (c+5) = 0 \)
This line passes through the center of C\(_1\), (0, -1). Substitute these coordinates:
\( 2g(0) + (2f-2)(-1) + (c+5) = 0 \)
\( -2f + 2 + c + 5 = 0 \implies c = 2f - 7 \) (Equation 1)
Tangency Condition for Circle C:
Circle C touches the line \(x - 2y = 0\) at the point T(2, 1).
Since C passes through T(2, 1), we substitute this point into its equation:
\( (2)^2 + (1)^2 + 2g(2) + 2f(1) + c = 0 \)
\( 5 + 4g + 2f + c = 0 \) (Equation 2)
The center of C, O, is \((-g, -f)\). The line joining the center O and the point of tangency T is perpendicular to the tangent line.
The slope of the tangent line \(x - 2y = 0\) is \(m_t = 1/2\).
The slope of the line OT is \(m_{OT} = \frac{-f - 1}{-g - 2}\).
Since the lines are perpendicular, \(m_t \cdot m_{OT} = -1\).
\( \frac{1}{2} \cdot \frac{-f - 1}{-g - 2} = -1 \)
\( -f - 1 = -2(-g - 2) = 2g + 4 \)
\( 2g + f = -5 \) (Equation 3)
Solving for g, f, and c:
From Equation 3, \( f = -5 - 2g \).
Substitute this into Equation 1: \( c = 2(-5 - 2g) - 7 = -10 - 4g - 7 = -17 - 4g \).
Now substitute f and c into Equation 2:
\( 5 + 4g + 2(-5 - 2g) + (-17 - 4g) = 0 \)
\( 5 + 4g - 10 - 4g - 17 - 4g = 0 \)
\( -22 - 4g = 0 \implies 4g = -22 \implies g = -11/2 \)
Now find f: \( f = -5 - 2(-11/2) = -5 + 11 = 6 \).
The center of C is \((-g, -f) = (11/2, -6)\).
Find c: \( c = 2f - 7 = 2(6) - 7 = 12 - 7 = 5 \).
The equation of circle C is \( x^2 + y^2 - 11x + 12y + 5 = 0 \).
Diameter of Circle C:
The radius of C, r, is given by \( r = \sqrt{g^2 + f^2 - c} = \sqrt{(-11/2)^2 + 6^2 - 5} \).
\( r^2 = \frac{121}{4} + 36 - 5 = \frac{121}{4} + 31 = \frac{121 + 124}{4} = \frac{245}{4} \)
\( r = \sqrt{\frac{245}{4}} = \frac{\sqrt{49 \times 5}}{2} = \frac{7\sqrt{5}}{2} \)
The diameter of C is \( 2r = 2 \times \frac{7\sqrt{5}}{2} = 7\sqrt{5} \).
Step 4: Final Answer:
The diameter of C is \(7\sqrt{5}\). This corresponds to option (D).