Circle with centre (-1,2) and passing through origin is

Options

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

Answer

To find the equation of the circle with center (-1, 2) that passes through the origin (0, 0), we can follow these steps:

### Step 1: Identify the center and the point on the circle
The center of the circle is given as \( C(-1, 2) \) and it passes through the origin \( P(0, 0) \).

### Step 2: Calculate the radius
The radius \( r \) of the circle is the distance from the center \( C \) to the point \( P \). We can use the distance formula:

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

Here, \( C(-1, 2) \) is \( (x_1, y_1) \) and \( P(0, 0) \) is \( (x_2, y_2) \).

Substituting the values:

\[ r = \sqrt{(0 - (-1))^2 + (0 - 2)^2} \]
\[ = \sqrt{(0 + 1)^2 + (-2)^2} \]
\[ = \sqrt{1^2 + 2^2} \]
\[ = \sqrt{1 + 4} \]
\[ = \sqrt{5} \]

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

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

Substituting \( h = -1 \), \( k = 2 \), and \( r = \sqrt{5} \):

\[ (x - (-1))^2 + (y - 2)^2 = (\sqrt{5})^2 \]
\[ (x + 1)^2 + (y - 2)^2 = 5 \]

### Step 4: Expand the equation
Now we expand the left side:

\[ (x + 1)^2 = x^2 + 2x + 1 \]
\[ (y - 2)^2 = y^2 - 4y + 4 \]

Combining these:

\[ x^2 + 2x + 1 + y^2 - 4y + 4 = 5 \]

### Step 5: Simplify the equation
Combine like terms:

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

Subtract 5 from both sides:

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

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

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