Find the equation of the tangent and
normal at `(1, 1)` to the circle `2x^(2) + 2 y^(2) - 2x - 5y + 3 = 0`

Answer

To find the equations of the tangent and normal at the point (1, 1) to the circle given by the equation \(2x^2 + 2y^2 - 2x - 5y + 3 = 0\), we will follow these steps:

### Step 1: Rewrite the Circle Equation
First, we can rewrite the equation of the circle in standard form. We start with:
\[ 2x^2 + 2y^2 - 2x - 5y + 3 = 0 \]
Dividing the entire equation by 2 gives:
\[ x^2 + y^2 - x - \frac{5}{2}y + \frac{3}{2} = 0 \]

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

For \(x\):
\[ x^2 - x = (x - \frac{1}{2})^2 - \frac{1}{4} \]

For \(y\):
\[ y^2 - \frac{5}{2}y = (y - \frac{5}{4})^2 - \frac{25}{16} \]

Substituting these back into the equation gives:
\[ (x - \frac{1}{2})^2 - \frac{1}{4} + (y - \frac{5}{4})^2 - \frac{25}{16} + \frac{3}{2} = 0 \]

Combining constants:
\[ -\frac{1}{4} - \frac{25}{16} + \frac{24}{16} = -\frac{1}{4} - \frac{1}{16} = -\frac{4}{16} - \frac{1}{16} = -\frac{5}{16} \]

So, we have:
\[ (x - \frac{1}{2})^2 + (y - \frac{5}{4})^2 = \frac{5}{16} \]

This represents a circle with center \((\frac{1}{2}, \frac{5}{4})\) and radius \(\frac{\sqrt{5}}{4}\).

### Step 3: Find the Slope of the Tangent
To find the slope of the tangent line at the point (1, 1), we differentiate the circle's equation implicitly:
\[ \frac{d}{dx}(2x^2 + 2y^2 - 2x - 5y + 3) = 0 \]
This gives:
\[ 4x + 4y\frac{dy}{dx} - 2 - 5\frac{dy}{dx} = 0 \]
Rearranging gives:
\[ (4y - 5)\frac{dy}{dx} = 2 - 4x \]
Thus, we have:
\[ \frac{dy}{dx} = \frac{2 - 4x}{4y - 5} \]

### Step 4: Evaluate the Derivative at (1, 1)
Substituting \(x = 1\) and \(y = 1\):
\[ \frac{dy}{dx} = \frac{2 - 4(1)}{4(1) - 5} = \frac{2 - 4}{4 - 5} = \frac{-2}{-1} = 2 \]

### Step 5: Write the Equation of the Tangent Line
Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
where \(m = 2\) and \((x_1, y_1) = (1, 1)\):
\[ y - 1 = 2(x - 1) \]
This simplifies to:
\[ y - 1 = 2x - 2 \implies y = 2x - 1 \]

### Step 6: Write the Equation of the Normal Line
The slope of the normal line is the negative reciprocal of the tangent slope:
\[ m_{\text{normal}} = -\frac{1}{2} \]
Using the point-slope form again:
\[ y - 1 = -\frac{1}{2}(x - 1) \]
This simplifies to:
\[ y - 1 = -\frac{1}{2}x + \frac{1}{2} \implies 2y - 2 = -x + 1 \implies x + 2y - 3 = 0 \]

### Final Answers
- The equation of the tangent line is \(y = 2x - 1\).
- The equation of the normal line is \(x + 2y - 3 = 0\).