Find the equation of the circle with centre `(-2, 3)` cutting a chord length 2 units
on ` 3x + 4y + 4 = 0 `

Answer

To find the equation of the circle with center \((-2, 3)\) that cuts a chord of length 2 units on the line \(3x + 4y + 4 = 0\), we can follow these steps:

### Step 1: Identify the center and the line equation
The center of the circle is given as \(C(-2, 3)\) and the line equation is \(3x + 4y + 4 = 0\).

### Step 2: Calculate the perpendicular distance from the center to the line
We can use the formula for the perpendicular distance \(D\) from a point \((x_0, y_0)\) to the line \(Ax + By + C = 0\):
\[ D = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For our line \(3x + 4y + 4 = 0\), we have \(A = 3\), \(B = 4\), and \(C = 4\). The coordinates of the center are \((-2, 3)\).

Substituting these values into the distance formula:
\[ D = \frac{|3(-2) + 4(3) + 4|}{\sqrt{3^2 + 4^2}} = \frac{|-6 + 12 + 4|}{\sqrt{9 + 16}} = \frac{|10|}{5} = 2 \]

### Step 3: Relate the distance to the chord length
The distance \(D\) from the center of the circle to the line is 2 units. The length of the chord \(L\) is given as 2 units. The relationship between the radius \(R\), the distance \(D\), and half the chord length \(\lambda\) is given by:
\[ R^2 = D^2 + \lambda^2 \]
Here, \(\lambda = \frac{L}{2} = \frac{2}{2} = 1\).

### Step 4: Calculate the radius
Substituting the known values into the formula:
\[ R^2 = 2^2 + 1^2 = 4 + 1 = 5 \]
Thus, \(R = \sqrt{5}\).

### Step 5: Write the equation of the circle
The standard form of the equation of a circle with center \((h, k)\) and radius \(R\) is:
\[ (x - h)^2 + (y - k)^2 = R^2 \]
Substituting \(h = -2\), \(k = 3\), and \(R^2 = 5\):
\[ (x + 2)^2 + (y - 3)^2 = 5 \]

### Step 6: Expand the equation
Expanding the equation:
\[ (x + 2)^2 + (y - 3)^2 = 5 \]
\[ x^2 + 4x + 4 + y^2 - 6y + 9 = 5 \]
Combining like terms:
\[ x^2 + y^2 + 4x - 6y + 8 = 0 \]

### Final Answer
The equation of the circle is:
\[ x^2 + y^2 + 4x - 6y + 8 = 0 \]

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