Find the equations of the tangents to the circle `x^(2) + y^(2) = 25 ` which are parallel to the line `2x- u + 4 = 0 `

Answer

To find the equations of the tangents to the circle \( x^2 + y^2 = 25 \) that are parallel to the line \( 2x - y + 4 = 0 \), we can follow these steps:

### Step 1: Identify the center and radius of the circle
The equation of the circle can be rewritten in standard form:
\[ x^2 + y^2 = 25 \implies (x - 0)^2 + (y - 0)^2 = 5^2 \]
From this, we can see that the center of the circle is \( (0, 0) \) and the radius \( r = 5 \).

### Step 2: Determine the slope of the given line
The equation of the line is given as \( 2x - y + 4 = 0 \). We can rearrange this into slope-intercept form \( y = mx + c \):
\[ y = 2x + 4 \]
From this, we can see that the slope \( m \) of the line is \( 2 \).

### Step 3: Write the equation of the tangent line
Since we want the tangents to be parallel to the line, the slope of the tangent lines will also be \( 2 \). Therefore, the equation of the tangent line can be written in the form:
\[ y = 2x + c \]
where \( c \) is a constant that we need to determine.

### Step 4: Use the distance formula
The distance from the center of the circle \( (0, 0) \) to the tangent line \( 2x - y + c = 0 \) must equal the radius of the circle. The distance \( d \) from a point \( (h, k) \) to a line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ah + Bk + C|}{\sqrt{A^2 + B^2}} \]
For our case:
- \( A = 2 \)
- \( B = -1 \)
- \( C = c \)
- \( h = 0 \)
- \( k = 0 \)

Substituting these values into the distance formula gives:
\[ d = \frac{|2(0) - 1(0) + c|}{\sqrt{2^2 + (-1)^2}} = \frac{|c|}{\sqrt{4 + 1}} = \frac{|c|}{\sqrt{5}} \]

### Step 5: Set the distance equal to the radius
Since the distance must equal the radius \( r = 5 \), we set up the equation:
\[ \frac{|c|}{\sqrt{5}} = 5 \]
Multiplying both sides by \( \sqrt{5} \) gives:
\[ |c| = 5\sqrt{5} \]

### Step 6: Solve for \( c \)
This implies that:
\[ c = 5\sqrt{5} \quad \text{or} \quad c = -5\sqrt{5} \]

### Step 7: Write the equations of the tangents
Substituting back into the tangent line equation \( y = 2x + c \), we have two equations:
1. \( y = 2x + 5\sqrt{5} \)
2. \( y = 2x - 5\sqrt{5} \)

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