Let PQ be the common chord of the circles `S_(1):x^(2)+y^(2)+2x+3y+1=0` and `S_(2):x^(2)+y^(2)+4x+3y+2=0`, then the perimeter (in units) of the triangle `C_(1)PQ` is equal to
`("where, "C_(1)=(-1, (-3)/(2)))`

Options

A. `(9)/(2)`
B. `2sqrt2+3`
C. `3sqrt2+3`
D. `(3)/(2)+2sqrt2`

Answer

To solve the problem, we need to find the perimeter of triangle \( C_1PQ \) where \( C_1 = (-1, -\frac{3}{2}) \) and \( PQ \) is the common chord of the circles given by the equations:

1. \( S_1: x^2 + y^2 + 2x + 3y + 1 = 0 \)
2. \( S_2: x^2 + y^2 + 4x + 3y + 2 = 0 \)

### Step 1: Find the centers and radii of the circles

The general form of a circle is given by \( x^2 + y^2 + 2gx + 2fy + c = 0 \), where the center is \( (-g, -f) \) and the radius \( r \) is given by \( r = \sqrt{g^2 + f^2 - c} \).

For \( S_1 \):
- \( g = 1 \), \( f = \frac{3}{2} \), \( c = 1 \)
- Center \( C_1 = (-1, -\frac{3}{2}) \)
- Radius \( r_1 = \sqrt{1^2 + \left(\frac{3}{2}\right)^2 - 1} = \sqrt{1 + \frac{9}{4} - 1} = \sqrt{\frac{9}{4}} = \frac{3}{2} \)

For \( S_2 \):
- \( g = 2 \), \( f = \frac{3}{2} \), \( c = 2 \)
- Center \( C_2 = (-2, -\frac{3}{2}) \)
- Radius \( r_2 = \sqrt{2^2 + \left(\frac{3}{2}\right)^2 - 2} = \sqrt{4 + \frac{9}{4} - 2} = \sqrt{\frac{13}{4}} = \frac{\sqrt{13}}{2} \)

### Step 2: Find the equation of the common chord \( PQ \)

The equation of the common chord can be found by subtracting the equations of the circles:

\[ S_1 - S_2 = 0 \implies (x^2 + y^2 + 2x + 3y + 1) - (x^2 + y^2 + 4x + 3y + 2) = 0 \]
This simplifies to:

\[ -2x - 1 = 0 \implies 2x + 1 = 0 \implies x = -\frac{1}{2} \]

### Step 3: Find the length of the common chord \( PQ \)

Since \( PQ \) is a vertical line at \( x = -\frac{1}{2} \), we need to find the intersection points with the circles.

Substituting \( x = -\frac{1}{2} \) into the equation of \( S_1 \):

\[ \left(-\frac{1}{2}\right)^2 + y^2 + 2\left(-\frac{1}{2}\right) + 3y + 1 = 0 \]
\[ \frac{1}{4} + y^2 - 1 + 3y + 1 = 0 \implies y^2 + 3y + \frac{1}{4} = 0 \]

Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ y = \frac{-3 \pm \sqrt{9 - 1}}{2} = \frac{-3 \pm \sqrt{8}}{2} = \frac{-3 \pm 2\sqrt{2}}{2} = -\frac{3}{2} \pm \sqrt{2} \]

The points \( P \) and \( Q \) are:

\[ P = \left(-\frac{1}{2}, -\frac{3}{2} + \sqrt{2}\right), \quad Q = \left(-\frac{1}{2}, -\frac{3}{2} - \sqrt{2}\right) \]

The length \( PQ \) is:

\[ PQ = \left| \left(-\frac{3}{2} + \sqrt{2}\right) - \left(-\frac{3}{2} - \sqrt{2}\right) \right| = 2\sqrt{2} \]

### Step 4: Calculate the perimeter of triangle \( C_1PQ \)

The perimeter \( P \) of triangle \( C_1PQ \) is given by:

\[ P = C_1P + C_1Q + PQ \]

Since \( C_1P = C_1Q = r_1 = \frac{3}{2} \):

\[ P = \frac{3}{2} + \frac{3}{2} + 2\sqrt{2} = 3 + 2\sqrt{2} \]

### Final Answer

The perimeter of triangle \( C_1PQ \) is \( 3 + 2\sqrt{2} \) units.