The equation of the circle having the lines `x^(2)+2xy+3x+6y=0` as its normals and having size just suffcient to contain the circle `x(x-4) + y(y - 3) = 0` is

Options

A. `x^(2)+y^(2)+3x-6y-40=0`
B. `x^(2)+y^(2)+6x-3y-45=0`
C. `x^(2)+y^(2)+8x+4y-20=0`
D. `x^(2)+y^(2)+4x+8y+20=0`

Answer

To find the equation of the circle that has the lines \(x^2 + 2xy + 3x + 6y = 0\) as its normals and is large enough to contain the circle given by \(x(x-4) + y(y-3) = 0\), we can follow these steps:

### Step 1: Identify the normals from the given equation
The equation of the normals is given as:
\[ x^2 + 2xy + 3x + 6y = 0 \]
We can factor this equation to find the lines. Rearranging gives:
\[ x^2 + 3x + 2xy + 6y = 0 \]
This can be factored as:
\[ x(x + 2y + 3) + 6y = 0 \]
From this, we can set \(x + 2y + 3 = 0\) and \(x + 3 = 0\).

### Step 2: Solve for the lines
From \(x + 2y + 3 = 0\):
\[ x + 2y = -3 \implies y = -\frac{1}{2}x - \frac{3}{2} \]
From \(x + 3 = 0\):
\[ x = -3 \]
Thus, the lines are:
1. \(x + 2y + 3 = 0\)
2. \(x + 3 = 0\)

### Step 3: Find the intersection point of the normals
To find the center of the circle, we need the intersection of the two lines:
Substituting \(x = -3\) into \(x + 2y + 3 = 0\):
\[ -3 + 2y + 3 = 0 \implies 2y = 0 \implies y = 0 \]
Thus, the center of the required circle is:
\[ C(-3, 0) \]

### Step 4: Find the radius of the given circle
The equation of the given circle is:
\[ x(x - 4) + y(y - 3) = 0 \]
This can be rewritten as:
\[ x^2 - 4x + y^2 - 3y = 0 \]
This represents a circle with diameter endpoints at \(A(0, 0)\) and \(B(4, 3)\). The center \(C_g\) of this circle is:
\[ C_g\left(\frac{0 + 4}{2}, \frac{0 + 3}{2}\right) = (2, 1.5) \]
The radius \(r\) is half the distance between points \(A\) and \(B\):
\[ AB = \sqrt{(4 - 0)^2 + (3 - 0)^2} = \sqrt{16 + 9} = 5 \]
Thus, the radius is:
\[ r = \frac{5}{2} = 2.5 \]

### Step 5: Determine the radius of the required circle
The distance \(d\) between the centers \(C(-3, 0)\) and \(C_g(2, 1.5)\) is:
\[ d = \sqrt{(-3 - 2)^2 + (0 - 1.5)^2} = \sqrt{(-5)^2 + (-1.5)^2} = \sqrt{25 + 2.25} = \sqrt{27.25} \]
This distance must equal the radius of the required circle \(R\) minus the radius of the given circle \(r\):
\[ d = R - r \]
Thus, we have:
\[ \sqrt{27.25} = R - 2.5 \]
Solving for \(R\):
\[ R = \sqrt{27.25} + 2.5 \]

### Step 6: Write the equation of the required circle
The equation of the circle with center \((-3, 0)\) and radius \(R\) is:
\[ (x + 3)^2 + (y - 0)^2 = R^2 \]
Substituting \(R\):
\[ (x + 3)^2 + y^2 = \left(\sqrt{27.25} + 2.5\right)^2 \]

### Final Equation
After calculating \(R^2\) and simplifying, we can express the equation in standard form.