Equation of the circle with centre at (1,-2) , and passing through the centre of the circle ` x^(2) + y^(2) + 2y - 3 = 0 ` , is

Options

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

Answer

To find the equation of the circle with center at (1, -2) and passing through the center of the circle defined by the equation \(x^2 + y^2 + 2y - 3 = 0\), we can follow these steps:

### Step 1: Find the center of the given circle
The given equation of the circle is:
\[ x^2 + y^2 + 2y - 3 = 0 \]
To find the center, we can rewrite this equation in the standard form. First, we complete the square for the \(y\) terms.

1. Rearranging the equation:
\[ x^2 + (y^2 + 2y) = 3 \]

2. Completing the square for \(y^2 + 2y\):
\[ y^2 + 2y = (y + 1)^2 - 1 \]
Thus, substituting back we have:
\[ x^2 + (y + 1)^2 - 1 = 3 \]
\[ x^2 + (y + 1)^2 = 4 \]

From this, we can see that the center of the circle is at \((0, -1)\).

### Step 2: Determine the radius of the new circle
The radius of the new circle can be calculated using the distance from its center \((1, -2)\) to the center of the given circle \((0, -1)\).

1. Using the distance formula:
\[ r = \sqrt{(1 - 0)^2 + (-2 - (-1))^2} \]
\[ = \sqrt{(1)^2 + (-2 + 1)^2} \]
\[ = \sqrt{1 + (-1)^2} \]
\[ = \sqrt{1 + 1} = \sqrt{2} \]

### Step 3: Write the equation of the circle
Now that we have the center \((1, -2)\) and the radius \(r = \sqrt{2}\), we can write the equation of the circle in standard form:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \(h = 1\), \(k = -2\), and \(r^2 = 2\).

Substituting these values:
\[ (x - 1)^2 + (y + 2)^2 = 2 \]

### Step 4: Expand the equation
Now we can expand this equation:
\[ (x - 1)^2 + (y + 2)^2 = 2 \]
Expanding both squares:
\[ (x^2 - 2x + 1) + (y^2 + 4y + 4) = 2 \]
Combining like terms:
\[ x^2 + y^2 - 2x + 4y + 5 = 2 \]
Subtracting 2 from both sides:
\[ x^2 + y^2 - 2x + 4y + 3 = 0 \]

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

---