The length of the common chord of the circles `x^(2)+y^(2)-2x-1=0` and `x^(2)+y^(2)+4y-1=0`, is

Options

A. `sqrt(15//2)`
B. `sqrt(15)`
C. `2sqrt(15)`
D. none of these

Answer

To find the length of the common chord of the circles given by the equations \( x^2 + y^2 - 2x - 1 = 0 \) and \( x^2 + y^2 + 4y - 1 = 0 \), we can follow these steps:

### Step 1: Rewrite the equations of the circles in standard form
The first circle's equation is:
\[ x^2 + y^2 - 2x - 1 = 0 \]
We can rearrange it to complete the square:
\[ (x^2 - 2x) + y^2 = 1 \]
Completing the square for \(x\):
\[ (x - 1)^2 - 1 + y^2 = 1 \implies (x - 1)^2 + y^2 = 2 \]
This represents a circle with center \((1, 0)\) and radius \(\sqrt{2}\).

The second circle's equation is:
\[ x^2 + y^2 + 4y - 1 = 0 \]
Rearranging and completing the square for \(y\):
\[ x^2 + (y^2 + 4y) = 1 \]
Completing the square for \(y\):
\[ x^2 + (y + 2)^2 - 4 = 1 \implies x^2 + (y + 2)^2 = 5 \]
This represents a circle with center \((0, -2)\) and radius \(\sqrt{5}\).

### Step 2: Find the distance between the centers of the circles
Let \(C_1 = (1, 0)\) and \(C_2 = (0, -2)\). The distance \(d\) between the centers is given by:
\[ d = \sqrt{(1 - 0)^2 + (0 - (-2))^2} = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \]

### Step 3: Use the formula for the length of the common chord
The length \(L\) of the common chord can be found using the formula:
\[ L = \sqrt{(r_1^2 - d^2) + (r_2^2 - d^2)} \]
where \(r_1\) and \(r_2\) are the radii of the circles, and \(d\) is the distance between the centers.

Here, \(r_1 = \sqrt{2}\) and \(r_2 = \sqrt{5}\).

Calculating \(r_1^2\) and \(r_2^2\):
\[ r_1^2 = 2, \quad r_2^2 = 5 \]
Now substituting into the formula:
\[ L = \sqrt{(2 - 5) + (5 - 5)} = \sqrt{(-3) + 0} = \sqrt{-3} \]
This indicates that the circles do not intersect, and hence there is no common chord.

### Conclusion
Since the circles do not intersect, the length of the common chord is \(0\).