Equation of a circle having radius equal to twice the radius of the circle `x^2+y^2+(2p +3)x + (3-2p)y +p-3 = 0` and touching it at the origin is

Options

A. `x^(2)+y^(2) +9x -3y = 0`
B. `x^(2) +y^(2) - 9x +3y = 0`
C. `x^(2) +y^(2) +18x +6y = 0`
D. `x^(2) +y^(2) +18x - 6y = 0`

Explanation

To find the equation of a circle that has a radius equal to twice the radius of the given circle and touches it at the origin, we will follow these steps:

### Step 1: Identify the given equation of the circle
The given equation is:
\[ x^2 + y^2 + (2p + 3)x + (3 - 2p)y + (p - 3) = 0 \]

### Step 2: Determine the center and radius of the given circle
The general form of a circle's equation is:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
From the given equation, we can identify:
- \( 2g = 2p + 3 \) → \( g = p + \frac{3}{2} \)
- \( 2f = 3 - 2p \) → \( f = \frac{3 - 2p}{2} \)
- \( c = p - 3 \)

The center of the circle is at \( (-g, -f) \) and the radius \( r \) is given by:
\[ r = \sqrt{g^2 + f^2 - c} \]

### Step 3: Substitute \( p \) to find the radius
Since the circles touch at the origin, we substitute \( (0, 0) \) into the equation:
\[ 0^2 + 0^2 + (2p + 3)(0) + (3 - 2p)(0) + (p - 3) = 0 \]
This simplifies to:
\[ p - 3 = 0 \]
Thus, \( p = 3 \).

### Step 4: Substitute \( p \) back into the circle's equation
Now, substitute \( p = 3 \) into the original equation:
\[ x^2 + y^2 + (2(3) + 3)x + (3 - 2(3))y + (3 - 3) = 0 \]
This simplifies to:
\[ x^2 + y^2 + 9x - 3y = 0 \]

### Step 5: Find the radius of the circle
Now, we can find the center and radius of this circle:
- \( g = 3 + \frac{3}{2} = \frac{9}{2} \)
- \( f = \frac{3 - 6}{2} = -\frac{3}{2} \)
- \( c = 0 \)

The radius \( r \) is:
\[ r = \sqrt{g^2 + f^2 - c} = \sqrt{\left(\frac{9}{2}\right)^2 + \left(-\frac{3}{2}\right)^2 - 0} \]
\[ r = \sqrt{\frac{81}{4} + \frac{9}{4}} = \sqrt{\frac{90}{4}} = \sqrt{\frac{45}{2}} \]

### Step 6: Find the equation of the new circle
The new circle has a radius equal to twice the radius of the original circle:
\[ R = 2r = 2\sqrt{\frac{90}{4}} = \sqrt{90} \]

The center of the new circle will be the same as the original circle, so:
\[ x + \frac{9}{2} \text{ and } y - \frac{3}{2} \]

The equation of the new circle is:
\[ \left(x + \frac{9}{2}\right)^2 + \left(y - \frac{3}{2}\right)^2 = 90 \]

### Step 7: Convert to general form
Expanding this gives:
\[ x^2 + 9x + \frac{81}{4} + y^2 - 3y + \frac{9}{4} = 90 \]
Combining terms:
\[ x^2 + y^2 + 9x - 3y + \frac{90}{4} - 90 = 0 \]
This simplifies to:
\[ x^2 + y^2 + 9x - 3y = 0 \]

### Final Answer
Thus, the equation of the circle is:
\[ x^2 + y^2 + 9x - 3y = 0 \]