Find the equation of the circle passing through the point ( 2, 3) and touching the line 2x+3y=4 at the point ( 2, 0)

Answer

To find the equation of the circle that passes through the point (2, 3) and touches the line \(2x + 3y = 4\) at the point (2, 0), we can follow these steps:

### Step 1: Identify the general form of the circle
The general equation of a circle can be expressed as:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \((h, k)\) is the center of the circle and \(r\) is the radius.

### Step 2: Use the point of tangency
Since the circle touches the line \(2x + 3y = 4\) at the point (2, 0), we can express the equation of the circle in the following form:
\[ (x - 2)^2 + (y - 0)^2 + \lambda(2x + 3y - 4) = 0 \]
where \(\lambda\) is a constant that we will determine.

### Step 3: Substitute the point (2, 3)
The circle must pass through the point (2, 3). We substitute \(x = 2\) and \(y = 3\) into the equation:
\[ (2 - 2)^2 + (3 - 0)^2 + \lambda(2(2) + 3(3) - 4) = 0 \]
This simplifies to:
\[ 0 + 9 + \lambda(4 + 9 - 4) = 0 \]
\[ 9 + 9\lambda = 0 \]

### Step 4: Solve for \(\lambda\)
From the equation \(9 + 9\lambda = 0\), we can solve for \(\lambda\):
\[ 9\lambda = -9 \implies \lambda = -1 \]

### Step 5: Substitute \(\lambda\) back into the circle equation
Now we substitute \(\lambda = -1\) back into the circle equation:
\[ (x - 2)^2 + y^2 - (2x + 3y - 4) = 0 \]
This simplifies to:
\[ (x - 2)^2 + y^2 - 2x - 3y + 4 = 0 \]

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

### Final Equation of the Circle
Thus, the equation of the circle is:
\[ x^2 + y^2 - 6x - 3y + 8 = 0 \]