Find the equation of the tangents to the circle `x^(2)+y^(2)-4x-5y+3=0` which are inclined at `45^(@)` with X axis.

Answer

To find the equations of the tangents to the circle given by the equation \( x^2 + y^2 - 4x - 5y + 3 = 0 \) that are inclined at \( 45^\circ \) with the x-axis, we can follow these steps:

### Step 1: Rewrite the Circle Equation
First, we need to rewrite the equation of the circle in standard form. We can do this by completing the square for both \( x \) and \( y \).

1. Rearranging the equation:
\[ x^2 - 4x + y^2 - 5y + 3 = 0 \]

2. Completing the square:
- For \( x^2 - 4x \):
\[ x^2 - 4x = (x - 2)^2 - 4 \]
- For \( y^2 - 5y \):
\[ y^2 - 5y = (y - \frac{5}{2})^2 - \frac{25}{4} \]

3. Substitute back into the equation:
\[ (x - 2)^2 - 4 + (y - \frac{5}{2})^2 - \frac{25}{4} + 3 = 0 \]
Simplifying gives:
\[ (x - 2)^2 + (y - \frac{5}{2})^2 = \frac{25}{4} - 4 + 3 = \frac{9}{4} \]

Thus, the center of the circle is \( (2, \frac{5}{2}) \) and the radius is \( \frac{3}{2} \).

### Step 2: Determine the Slope of the Tangents
Since the tangents are inclined at \( 45^\circ \) with the x-axis, the slope \( m \) of the tangent lines is:
\[ m = \tan(45^\circ) = 1 \]
Thus, the equations of the tangents can be expressed as:
\[ y = x + c \]

### Step 3: Substitute into the Circle Equation
To find the value of \( c \), we substitute \( y = x + c \) into the circle's equation:
\[ x^2 + (x + c)^2 - 4x - 5(x + c) + 3 = 0 \]

Expanding this:
\[ x^2 + (x^2 + 2xc + c^2) - 4x - 5x - 5c + 3 = 0 \]
This simplifies to:
\[ 2x^2 + (2c - 9)x + (c^2 - 5c + 3) = 0 \]

### Step 4: Condition for Tangency
For the line to be a tangent to the circle, the quadratic equation must have real and equal roots. Therefore, the discriminant must be zero:
\[ b^2 - 4ac = 0 \]
Here, \( a = 2 \), \( b = 2c - 9 \), and \( c = c^2 - 5c + 3 \).

Setting the discriminant to zero:
\[ (2c - 9)^2 - 4 \cdot 2 \cdot (c^2 - 5c + 3) = 0 \]

### Step 5: Solve for \( c \)
Expanding and simplifying:
\[ 4c^2 - 36c + 81 - 8(c^2 - 5c + 3) = 0 \]
\[ 4c^2 - 36c + 81 - 8c^2 + 40c - 24 = 0 \]
Combining like terms:
\[ -4c^2 + 4c + 57 = 0 \]
Multiplying through by -1:
\[ 4c^2 - 4c - 57 = 0 \]

Using the quadratic formula:
\[ c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 4 \cdot (-57)}}{2 \cdot 4} \]
Calculating the discriminant:
\[ = \frac{4 \pm \sqrt{16 + 912}}{8} = \frac{4 \pm \sqrt{928}}{8} = \frac{4 \pm 4\sqrt{58}}{8} = \frac{1 \pm \sqrt{58}}{2} \]

### Step 6: Write the Tangent Equations
Substituting the values of \( c \) back into the tangent equations:
1. \( y = x + \frac{1 + \sqrt{58}}{2} \)
2. \( y = x + \frac{1 - \sqrt{58}}{2} \)

### Final Answer
The equations of the tangents to the circle that are inclined at \( 45^\circ \) with the x-axis are:
\[ y = x + \frac{1 + \sqrt{58}}{2} \quad \text{and} \quad y = x + \frac{1 - \sqrt{58}}{2} \]