Find the equation of the tangent to the circle `x^2+y^2-2x-2y-23=0` and parallel to `2x+y+3=0.`

Answer

To find the equation of the tangent to the circle given by the equation \( x^2 + y^2 - 2x - 2y - 23 = 0 \) that is parallel to the line \( 2x + y + 3 = 0 \), we can follow these steps:

### Step 1: Rewrite the Circle Equation
First, we rewrite the equation of the circle in standard form. The given equation is:
\[ x^2 + y^2 - 2x - 2y - 23 = 0 \]
We can rearrange it as:
\[ x^2 - 2x + y^2 - 2y = 23 \]
Next, we complete the square for both \( x \) and \( y \).

### Step 2: Complete the Square
For \( x^2 - 2x \):
\[ x^2 - 2x = (x - 1)^2 - 1 \]
For \( y^2 - 2y \):
\[ y^2 - 2y = (y - 1)^2 - 1 \]
Substituting back, we have:
\[ (x - 1)^2 - 1 + (y - 1)^2 - 1 = 23 \]
This simplifies to:
\[ (x - 1)^2 + (y - 1)^2 = 25 \]
Thus, the center of the circle is \( (1, 1) \) and the radius \( r = 5 \).

### Step 3: Identify the Slope of the Given Line
The line given is \( 2x + y + 3 = 0 \). We can rewrite this in slope-intercept form:
\[ y = -2x - 3 \]
From this, we see that the slope \( m \) of the line is \( -2 \).

### Step 4: Use the Tangent Equation
The equation of the tangent to the circle at a point with slope \( m \) can be expressed as:
\[ y - k = m(x - h) \pm r\sqrt{1 + m^2} \]
where \( (h, k) \) is the center of the circle, \( r \) is the radius, and \( m \) is the slope of the tangent.

Substituting \( h = 1 \), \( k = 1 \), \( r = 5 \), and \( m = -2 \):
\[ y - 1 = -2(x - 1) \pm 5\sqrt{1 + (-2)^2} \]
Calculating \( \sqrt{1 + 4} = \sqrt{5} \):
\[ y - 1 = -2(x - 1) \pm 5\sqrt{5} \]

### Step 5: Simplify the Tangent Equation
Now, we can simplify the equation:
\[ y - 1 = -2x + 2 \pm 5\sqrt{5} \]
Thus, we have two equations:
1. \( y = -2x + 3 + 5\sqrt{5} \)
2. \( y = -2x + 3 - 5\sqrt{5} \)

### Final Answer
The equations of the tangents to the circle that are parallel to the line \( 2x + y + 3 = 0 \) are:
\[ y = -2x + 3 + 5\sqrt{5} \]
and
\[ y = -2x + 3 - 5\sqrt{5} \]