Equation of normal to the circle `2x^(2)+2y^(2)+3x-4y+1=0` at (-1, 2) is ………..

Answer

To find the equation of the normal to the circle given by the equation \(2x^2 + 2y^2 + 3x - 4y + 1 = 0\) at the point \((-1, 2)\), we will follow these steps:

### Step 1: Rewrite the equation of the circle
First, we simplify the equation of the circle by dividing everything by 2:

\[ x^2 + y^2 + \frac{3}{2}x - 2y + \frac{1}{2} = 0 \]

### Step 2: Identify the coefficients
In the standard form of the circle's equation \(x^2 + y^2 + 2gx + 2fy + c = 0\), we can identify:

- \(g = \frac{3}{4}\)
- \(f = -1\)
- \(c = \frac{1}{2}\)

### Step 3: Use the formula for the normal
The formula for the equation of the normal to the circle at the point \((x_1, y_1)\) is given by:

\[ \frac{x - x_1}{x_1 + g} = \frac{y - y_1}{y_1 + f} \]

Substituting \(x_1 = -1\), \(y_1 = 2\), \(g = \frac{3}{4}\), and \(f = -1\):

\[ \frac{x + 1}{-1 + \frac{3}{4}} = \frac{y - 2}{2 - 1} \]

### Step 4: Simplify the left side
Calculate \(-1 + \frac{3}{4}\):

\[ -1 + \frac{3}{4} = -\frac{4}{4} + \frac{3}{4} = -\frac{1}{4} \]

Thus, the equation becomes:

\[ \frac{x + 1}{-\frac{1}{4}} = y - 2 \]

### Step 5: Cross-multiply to eliminate the fraction
Cross-multiplying gives:

\[ 4(x + 1) = -(y - 2) \]

### Step 6: Rearrange the equation
Expanding and rearranging:

\[ 4x + 4 = -y + 2 \]

Rearranging gives:

\[ 4x + y + 2 = 0 \]

### Final Answer
The equation of the normal to the circle at the point \((-1, 2)\) is:

\[ 4x + y + 2 = 0 \]