Find the distance between the centres of the circles `x^(2)-y^(2)+8x+10y-2=0andx^(2)-y^(2)+2x+2y-1=0`.

Answer

To find the distance between the centers of the given circles, we will follow these steps:

1. **Rewrite the equations of the circles**:
The given equations are:
\[ x^2 - y^2 + 8x + 10y - 2 = 0 \]
\[ x^2 - y^2 + 2x + 2y - 1 = 0 \]

2. **Identify the center of the first circle**:
The general form of a circle is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
To find the center, we can rewrite the first equation in the form of a circle. We can rearrange the equation:
\[ x^2 + 8x - y^2 + 10y - 2 = 0 \]
Completing the square for \(x\) and \(y\):
- For \(x^2 + 8x\):
\[ x^2 + 8x = (x + 4)^2 - 16 \]
- For \(-y^2 + 10y\):
\[ -y^2 + 10y = -(y^2 - 10y) = -(y - 5)^2 + 25 \]
Putting it all together:
\[ (x + 4)^2 - 16 - (y - 5)^2 + 25 - 2 = 0 \]
Simplifying:
\[ (x + 4)^2 - (y - 5)^2 + 7 = 0 \]
Thus, the center of the first circle \(C_1\) is \((-4, 5)\).

3. **Identify the center of the second circle**:
Now, we repeat the process for the second circle:
\[ x^2 + 2x - y^2 + 2y - 1 = 0 \]
Completing the square:
- For \(x^2 + 2x\):
\[ x^2 + 2x = (x + 1)^2 - 1 \]
- For \(-y^2 + 2y\):
\[ -y^2 + 2y = -(y^2 - 2y) = -(y - 1)^2 + 1 \]
Putting it all together:
\[ (x + 1)^2 - 1 - (y - 1)^2 + 1 - 1 = 0 \]
Simplifying:
\[ (x + 1)^2 - (y - 1)^2 - 1 = 0 \]
Thus, the center of the second circle \(C_2\) is \((-1, 1)\).

4. **Calculate the distance between the centers**:
The distance \(d\) between the two centers \(C_1(-4, 5)\) and \(C_2(-1, 1)\) can be calculated using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates:
\[ d = \sqrt{((-1) - (-4))^2 + (1 - 5)^2} \]
Simplifying:
\[ d = \sqrt{(3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

5. **Final Answer**:
The distance between the centers of the two circles is \(5\) units.