The equation of the circle passing through (3,6) and whose centre (2,-2) is

Options

A. `x^2+y^2-4x+2y=45`
B. `x^2+y^2-4x+4y=57`
C. `x^2+y^2+4x-2y=45`
D. `x^2+y^2-4x+2y+45=0`

Answer

To find the equation of the circle that passes through the point (3,6) and has its center at (2,-2), we will follow these steps:

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

### Step 2: Substitute the center (2, -2)
Here, the center of the circle is given as (2, -2). Thus, we substitute a = 2 and b = -2 into the general equation:
\[ (x - 2)^2 + (y + 2)^2 = r^2 \]

### Step 3: Substitute the point (3, 6)
Since the circle passes through the point (3, 6), we substitute x = 3 and y = 6 into the equation to find r:
\[ (3 - 2)^2 + (6 + 2)^2 = r^2 \]
Calculating this gives:
\[ (1)^2 + (8)^2 = r^2 \]
\[ 1 + 64 = r^2 \]
\[ r^2 = 65 \]

### Step 4: Write the equation of the circle
Now that we have r², we can write the equation of the circle:
\[ (x - 2)^2 + (y + 2)^2 = 65 \]

### Step 5: Expand the equation
Next, we will expand the equation:
\[ (x - 2)^2 + (y + 2)^2 = 65 \]
Expanding gives:
\[ (x^2 - 4x + 4) + (y^2 + 4y + 4) = 65 \]
Combining like terms:
\[ x^2 + y^2 - 4x + 4y + 8 = 65 \]
Subtracting 65 from both sides:
\[ x^2 + y^2 - 4x + 4y + 8 - 65 = 0 \]
This simplifies to:
\[ x^2 + y^2 - 4x + 4y - 57 = 0 \]

### Final Equation
Thus, the equation of the circle is:
\[ x^2 + y^2 - 4x + 4y = 57 \]