To the circle `x^(2)+y^(2)+8x-4y+4=0` tangent at the point `theta=(pi)/4` is

Options

A. a) `x+y+2-4sqrt(2)=0`
B. b) `x-y+2-4sqrt(2)=0`
C. c) `x+y+4+4sqrt(2)=0`
D. d) `x-y-2-4sqrt(2)=0`

Answer

To find the equation of the tangent to the circle given by the equation \(x^2 + y^2 + 8x - 4y + 4 = 0\) at the point corresponding to the angle \(\theta = \frac{\pi}{4}\), we will follow these steps:

### Step 1: Rewrite the Circle's Equation
First, we need to rewrite the equation of the circle in standard form. The given equation is:
\[ x^2 + y^2 + 8x - 4y + 4 = 0 \]
We can rearrange it to:
\[ x^2 + 8x + y^2 - 4y + 4 = 0 \]

### Step 2: Complete the Square
Next, we complete the square for both \(x\) and \(y\).

For \(x\):
\[ x^2 + 8x = (x + 4)^2 - 16 \]

For \(y\):
\[ y^2 - 4y = (y - 2)^2 - 4 \]

Substituting these back into the equation gives:
\[ (x + 4)^2 - 16 + (y - 2)^2 - 4 + 4 = 0 \]
Simplifying this, we have:
\[ (x + 4)^2 + (y - 2)^2 - 16 = 0 \]
Thus, the equation of the circle is:
\[ (x + 4)^2 + (y - 2)^2 = 16 \]
This shows that the center of the circle is at \((-4, 2)\) and the radius \(r = 4\).

### Step 3: Find the Point on the Circle
The point on the circle corresponding to \(\theta = \frac{\pi}{4}\) can be found using the parametric equations:
\[ x = -4 + r \cos(\theta) \]
\[ y = 2 + r \sin(\theta) \]
Substituting \(r = 4\) and \(\theta = \frac{\pi}{4}\):
\[ x = -4 + 4 \cos\left(\frac{\pi}{4}\right) = -4 + 4 \cdot \frac{1}{\sqrt{2}} = -4 + 2\sqrt{2} \]
\[ y = 2 + 4 \sin\left(\frac{\pi}{4}\right) = 2 + 4 \cdot \frac{1}{\sqrt{2}} = 2 + 2\sqrt{2} \]
Thus, the coordinates of the point on the circle are:
\[ \left(-4 + 2\sqrt{2}, 2 + 2\sqrt{2}\right) \]

### Step 4: Equation of the Tangent Line
The equation of the tangent line at the point \((x_1, y_1)\) on the circle can be expressed as:
\[ (x + 4)(x_1 + 4) + (y - 2)(y_1 - 2) = 16 \]
Substituting \(x_1 = -4 + 2\sqrt{2}\) and \(y_1 = 2 + 2\sqrt{2}\):
\[ (x + 4)(2\sqrt{2}) + (y - 2)(2\sqrt{2}) = 16 \]
Simplifying this:
\[ 2\sqrt{2}(x + 4 + y - 2) = 16 \]
\[ x + y + 2\sqrt{2} - 2 = 0 \]
Thus, the equation of the tangent line is:
\[ x + y + 2 - 4\sqrt{2} = 0 \]

### Final Answer
The equation of the tangent to the circle at the point corresponding to \(\theta = \frac{\pi}{4}\) is:
\[ x + y + 2 - 4\sqrt{2} = 0 \]