Find the equation of th circle whose centre is ` (2, -3)` and which passes through the intersection of the lines ` 3x + 2y = 11 and 2x + 3y = 4`.

Answer

To find the equation of the circle whose center is at (2, -3) and which passes through the intersection of the lines \(3x + 2y = 11\) and \(2x + 3y = 4\), we will follow these steps:

### Step 1: Find the intersection point of the two lines
We have the equations of the lines:
1. \(3x + 2y = 11\) (Equation 1)
2. \(2x + 3y = 4\) (Equation 2)

We can solve these equations simultaneously to find the intersection point.

#### Substituting for \(x\) from Equation 1:
From Equation 1, we can express \(x\) in terms of \(y\):
\[ 3x = 11 - 2y \implies x = \frac{11 - 2y}{3} \]

#### Substitute \(x\) in Equation 2:
Now, substitute this expression for \(x\) into Equation 2:
\[ 2\left(\frac{11 - 2y}{3}\right) + 3y = 4 \]
Multiply through by 3 to eliminate the fraction:
\[ 2(11 - 2y) + 9y = 12 \]
Expanding gives:
\[ 22 - 4y + 9y = 12 \]
Combine like terms:
\[ 22 + 5y = 12 \]
Now, isolate \(y\):
\[ 5y = 12 - 22 \implies 5y = -10 \implies y = -2 \]

#### Finding \(x\):
Substituting \(y = -2\) back into the expression for \(x\):
\[ x = \frac{11 - 2(-2)}{3} = \frac{11 + 4}{3} = \frac{15}{3} = 5 \]

Thus, the intersection point is \((5, -2)\).

### Step 2: Use the center and the point on the circumference to find the radius
The center of the circle is \((2, -3)\) and the point on the circumference is \((5, -2)\).

Using the distance formula to find the radius \(r\):
\[ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Where \((x_1, y_1) = (2, -3)\) and \((x_2, y_2) = (5, -2)\):
\[ r = \sqrt{(5 - 2)^2 + (-2 + 3)^2} = \sqrt{(3)^2 + (1)^2} = \sqrt{9 + 1} = \sqrt{10} \]

### Step 3: Write the equation of the circle
The general equation of a circle with center \((h, k)\) and radius \(r\) is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \(h = 2\), \(k = -3\), and \(r = \sqrt{10}\):
\[ (x - 2)^2 + (y + 3)^2 = (\sqrt{10})^2 \]
This simplifies to:
\[ (x - 2)^2 + (y + 3)^2 = 10 \]

### Step 4: Expand the equation to general form
Expanding the equation:
\[ (x - 2)^2 + (y + 3)^2 = 10 \]
\[ x^2 - 4x + 4 + y^2 + 6y + 9 = 10 \]
Combining like terms:
\[ x^2 + y^2 - 4x + 6y + 3 = 0 \]

### Final Answer
The equation of the circle is:
\[ x^2 + y^2 - 4x + 6y + 3 = 0 \]

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