Show that the tangent at `(-1, 2)` of the circle `x^(2) + y^(2) - 4x -8y + 7 = 0` touches the circle `x^(2) + y^(2) + 4x + 6y = 0` and also find its point of contact.

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the equation of the first circle
The first circle is given by the equation:
\[ x^2 + y^2 - 4x - 8y + 7 = 0 \]

### Step 2: Identify the center and radius of the first circle
We can rewrite the equation in standard form by completing the square.

1. For \(x\):
\[ x^2 - 4x = (x - 2)^2 - 4 \]
2. For \(y\):
\[ y^2 - 8y = (y - 4)^2 - 16 \]

Putting it all together:
\[ (x - 2)^2 - 4 + (y - 4)^2 - 16 + 7 = 0 \]
\[ (x - 2)^2 + (y - 4)^2 - 13 = 0 \]
Thus, the center is \((2, 4)\) and the radius is \(\sqrt{13}\).

### Step 3: Find the equation of the tangent at the point \((-1, 2)\)
The formula for the tangent to a circle at a point \((x_1, y_1)\) is:
\[ xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0 \]
where \(g = -2\) and \(f = -4\) from the circle's equation.

Substituting \(x_1 = -1\) and \(y_1 = 2\):
\[ x(-1) + y(2) - 2(x - 1) - 4(y + 2) + 7 = 0 \]

Simplifying:
\[ -x + 2y - 2x + 2 - 4y - 8 + 7 = 0 \]
\[ -3x - 2y + 1 = 0 \]
Rearranging gives:
\[ 3x + 2y - 1 = 0 \]

### Step 4: Verify if this tangent touches the second circle
The second circle is given by:
\[ x^2 + y^2 + 4x + 6y = 0 \]

Rearranging gives:
\[ x^2 + y^2 + 4x + 6y = 0 \]

### Step 5: Substitute \(y\) from the tangent equation into the second circle's equation
From the tangent equation \(3x + 2y - 1 = 0\), we can express \(y\):
\[ y = \frac{1 - 3x}{2} \]

Substituting into the second circle's equation:
\[ x^2 + \left(\frac{1 - 3x}{2}\right)^2 + 4x + 6\left(\frac{1 - 3x}{2}\right) = 0 \]

Expanding:
\[ x^2 + \frac{(1 - 3x)^2}{4} + 4x + 3(1 - 3x) = 0 \]
\[ x^2 + \frac{1 - 6x + 9x^2}{4} + 4x + 3 - 9x = 0 \]
Multiplying through by 4 to eliminate the fraction:
\[ 4x^2 + 1 - 6x + 9x^2 + 16x + 12 - 36x = 0 \]
Combining like terms:
\[ 13x^2 - 26x + 13 = 0 \]

### Step 6: Solve the quadratic equation
Dividing through by 13:
\[ x^2 - 2x + 1 = 0 \]
Factoring gives:
\[ (x - 1)^2 = 0 \]
Thus, \(x = 1\).

### Step 7: Find \(y\) corresponding to \(x = 1\)
Substituting \(x = 1\) back into the tangent equation:
\[ y = \frac{1 - 3(1)}{2} = \frac{1 - 3}{2} = -1 \]

### Conclusion
The point of contact where the tangent touches the second circle is:
\[ (1, -1) \]