The equation of the circle passing through (4, 5) having the centre (2, 2), is

Options

A. `x^(2)+y^(2)+4x+4y-5=0`
B. `x^(2)+y^(2)-4x-4y-5=0`
C. `x^(2)+y^(2)-4x=13`
D. `x^(2)+y^(2)-4x-4y+5=0`

Explanation

To find the equation of the circle passing through the point (4, 5) with the 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 (2, 2) and the point through which the circle passes is (4, 5).

### Step 2: Calculate the radius
The radius \( r \) of the circle can be calculated using the distance formula between the center and the point on the circle. The formula for the distance \( r \) is:

\[ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Substituting the coordinates of the center (2, 2) and the point (4, 5):

\[ r = \sqrt{(4 - 2)^2 + (5 - 2)^2} \]
\[ r = \sqrt{(2)^2 + (3)^2} \]
\[ r = \sqrt{4 + 9} \]
\[ r = \sqrt{13} \]

### Step 3: Write the equation of the circle
The general equation of a circle with center \((h, k)\) and radius \(r\) is given by:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

Here, \(h = 2\), \(k = 2\), and \(r = \sqrt{13}\). Therefore, substituting these values into the equation:

\[ (x - 2)^2 + (y - 2)^2 = (\sqrt{13})^2 \]
\[ (x - 2)^2 + (y - 2)^2 = 13 \]

### Step 4: Expand the equation
Now, we can expand the equation:

\[ (x - 2)^2 + (y - 2)^2 = 13 \]
\[ (x^2 - 4x + 4) + (y^2 - 4y + 4) = 13 \]
\[ x^2 - 4x + y^2 - 4y + 8 = 13 \]
\[ x^2 + y^2 - 4x - 4y + 8 - 13 = 0 \]
\[ x^2 + y^2 - 4x - 4y - 5 = 0 \]

### Final Equation
Thus, the equation of the circle is:

\[ x^2 + y^2 - 4x - 4y - 5 = 0 \]

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