Find the equation of the circle with centre (2,3) and passing through the point (2,-1).

Answer

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

### Step 1: Identify the center and the point
The center of the circle is given as (h, k) = (2, 3), and the point through which the circle passes is (x1, y1) = (2, -1).

### 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:

\[ r = \sqrt{(h - x1)^2 + (k - y1)^2} \]

Substituting the values:

\[ r = \sqrt{(2 - 2)^2 + (3 - (-1))^2} \]
\[ r = \sqrt{(0)^2 + (3 + 1)^2} \]
\[ r = \sqrt{0 + 4^2} = \sqrt{16} = 4 \]

### Step 3: Write the standard 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 the values of h, k, and r:

\[ (x - 2)^2 + (y - 3)^2 = 4^2 \]
\[ (x - 2)^2 + (y - 3)^2 = 16 \]

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

\[ (x - 2)^2 + (y - 3)^2 = 16 \]

Expanding both squares:

\[ (x^2 - 4x + 4) + (y^2 - 6y + 9) = 16 \]

Combining the terms:

\[ x^2 - 4x + y^2 - 6y + 13 = 16 \]

### Step 5: Rearranging the equation
Now, we will rearrange the equation to set it to zero:

\[ x^2 + y^2 - 4x - 6y + 13 - 16 = 0 \]
\[ x^2 + y^2 - 4x - 6y - 3 = 0 \]

### Final Answer
The equation of the circle is:

\[ x^2 + y^2 - 4x - 6y - 3 = 0 \]

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