Equation of circle centred at (2,-3) , and passing through (-1,2) , is

Options

A. `x^(2) + y^(2) - 4x - 6y - 34 = 0 `
B. `X^(2) + y^(2) + 4x - 6y + 34 = 0 `
C. `X^(2) + y^(2) + 4x + 6y+ 21 = 0 `
D. `X^(2) + y^(2) - 4x + 6y-21 = 0 `

Answer

To find the equation of the circle centered at (2, -3) and passing through the point (-1, 2), we can follow these steps:

### Step 1: Write the standard form of the circle's equation
The standard form of the equation of a circle is given by:
\[ (x - a)^2 + (y - b)^2 = r^2 \]
where \((a, b)\) is the center of the circle and \(r\) is the radius.

### Step 2: Substitute the center coordinates
Given the center of the circle is (2, -3), we substitute \(a = 2\) and \(b = -3\):
\[ (x - 2)^2 + (y + 3)^2 = r^2 \]

### Step 3: Find the radius using the point on the circle
The circle passes through the point (-1, 2). We can use this point to find the radius \(r\). Substitute \(x = -1\) and \(y = 2\) into the equation:
\[ (-1 - 2)^2 + (2 + 3)^2 = r^2 \]
Calculating this gives:
\[ (-3)^2 + (5)^2 = r^2 \]
\[ 9 + 25 = r^2 \]
\[ r^2 = 34 \]

### Step 4: Write the final equation of the circle
Now we can substitute \(r^2\) back into the equation:
\[ (x - 2)^2 + (y + 3)^2 = 34 \]

### Step 5: Expand the equation (optional)
If we want to expand this equation, we can do so:
\[ (x - 2)^2 = x^2 - 4x + 4 \]
\[ (y + 3)^2 = y^2 + 6y + 9 \]
Adding these together gives:
\[ x^2 - 4x + 4 + y^2 + 6y + 9 = 34 \]
Combining like terms:
\[ x^2 + y^2 - 4x + 6y + 13 - 34 = 0 \]
This simplifies to:
\[ x^2 + y^2 - 4x + 6y - 21 = 0 \]

### Final Answer
The equation of the circle is:
\[ (x - 2)^2 + (y + 3)^2 = 34 \]
or in expanded form:
\[ x^2 + y^2 - 4x + 6y - 21 = 0 \]