If the line `x=2y` touches circle C at (2,1) & C cuts `C_1=x^2+y^2+2y-5=0` such that common chord is diameter of `C_1`. Find diameter of C.

Answer

To solve the problem, we need to find the diameter of circle C given that it touches the line \( x = 2y \) at the point (2, 1) and intersects circle \( C_1 \) such that the common chord is the diameter of \( C_1 \).

### Step-by-Step Solution:

1. **Identify the Circle \( C_1 \)**:
The equation of circle \( C_1 \) is given as:
\[ x^2 + y^2 + 2y - 5 = 0 \]
We can rewrite this equation in standard form by completing the square for \( y \):
\[ x^2 + (y^2 + 2y + 1) - 1 - 5 = 0 \implies x^2 + (y + 1)^2 = 6 \]
Thus, the center of circle \( C_1 \) is \( (0, -1) \) and its radius is \( \sqrt{6} \).

2. **Equation of Line**:
The line \( x = 2y \) can be rewritten as:
\[ x - 2y = 0 \]

3. **Equation of Circle \( C \)**:
Since the line touches circle \( C \) at the point (2, 1), we can use the family of circles concept. The equation of circle \( C \) can be expressed as:
\[ (x - 2)^2 + (y - 1)^2 + \lambda (x - 2y) = 0 \]
Expanding this gives:
\[ (x - 2)^2 + (y - 1)^2 + \lambda (x - 2y) = 0 \]
\[ x^2 - 4x + 4 + y^2 - 2y + 1 + \lambda x - 2\lambda y = 0 \]
\[ x^2 + y^2 + (\lambda - 4)x + (-2\lambda - 2)y + 5 = 0 \]

4. **Common Chord Condition**:
The common chord of circles \( C \) and \( C_1 \) is a diameter of \( C_1 \). The equation for the common chord can be derived from:
\[ C - C_1 = 0 \]
Substituting the equations gives:
\[ \lambda x - 2\lambda y + 5 - (x^2 + (y + 1)^2 - 6) = 0 \]
Simplifying this leads to:
\[ \lambda x - 2\lambda y + 5 - x^2 - (y^2 + 2y + 1 - 6) = 0 \]
\[ \lambda x - 2\lambda y + 5 - x^2 - y^2 - 2y + 5 = 0 \]
\[ -x^2 - y^2 + \lambda x - (2\lambda + 2)y = 0 \]

5. **Substituting the Center of \( C_1 \)**:
Since the common chord is a diameter, it passes through the center of \( C_1 \) at \( (0, -1) \):
\[ \lambda(0) - 2\lambda(-1) + 10 = 0 \implies 2\lambda + 10 = 0 \implies \lambda = -5 \]

6. **Finding the Equation of Circle \( C \)**:
Substitute \( \lambda = -5 \) into the equation of circle \( C \):
\[ x^2 + y^2 - 9x + 8y + 5 = 0 \]

7. **Finding the Center and Radius of Circle \( C \)**:
The standard form of the circle is:
\[ (x - \frac{9}{2})^2 + (y + 4)^2 = r^2 \]
The center is \( \left(\frac{9}{2}, -4\right) \) and we can find the radius using:
\[ r^2 = \left(\frac{9}{2}\right)^2 + (-4)^2 - 5 = \frac{81}{4} + 16 - 5 = \frac{81 + 64 - 20}{4} = \frac{125}{4} \]
Thus, the radius \( r = \frac{5\sqrt{5}}{2} \).

8. **Finding the Diameter**:
The diameter \( D \) of circle \( C \) is:
\[ D = 2r = 2 \times \frac{5\sqrt{5}}{2} = 5\sqrt{5} \]

### Final Answer:
The diameter of circle \( C \) is \( 5\sqrt{5} \).