The equation of the circle which passes through the point (3, 4) and has its centre at (2, 2) is

Options

A. `x^(2) + y^(2) + 4x + 4y -1 =0`
B. `x^(2) + y^(2) - 4x - 4y = 2`
C. `x^(2) + y^(2) - 4x - 4y -1 =0`
D. `x^(2) + y^(2) - 4x - 4y + 3 = 0`

Answer

To find the equation of the circle that passes through the point (3, 4) and has its center at (2, 2), we can follow these steps:

### Step 1: Identify the center and the point on the circle
The center of the circle is given as \( (h, k) = (2, 2) \) and the point through which the circle passes is \( (x_1, y_1) = (3, 4) \).

### Step 2: Calculate the radius using the distance formula
The radius \( r \) of the circle can be calculated using the distance formula:
\[ r = \sqrt{(x_1 - h)^2 + (y_1 - k)^2} \]
Substituting the values:
\[ r = \sqrt{(3 - 2)^2 + (4 - 2)^2} = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \]

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

### Step 4: Expand the equation
Now, we will expand the equation:
\[ (x - 2)^2 + (y - 2)^2 = 5 \]
Expanding each term:
\[ (x^2 - 4x + 4) + (y^2 - 4y + 4) = 5 \]
Combining like terms:
\[ x^2 + y^2 - 4x - 4y + 8 = 5 \]
Rearranging gives:
\[ x^2 + y^2 - 4x - 4y + 3 = 0 \]

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

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