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 find the perimeter of triangle \( C_1PQ \), where \( C_1 = (-1, -\frac{3}{2}) \) is the center of the first circle and \( PQ \) is the common chord of the two circles given by the equations:

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

we will follow these steps:

### Step 1: Find the equation of the common chord \( PQ \)

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

\[ S_1 - S_2 = 0 \]

This gives us:

\[ (x^2 + y^2 + 2x + 3y + 1) - (x^2 + y^2 + 4x + 3y + 2) = 0 \]

Simplifying this:

\[ 2x + 3y + 1 - 4x - 3y - 2 = 0 \]

This simplifies to:

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

### Step 2: Substitute \( x = -\frac{1}{2} \) into one of the circle equations to find \( y \)

We can substitute \( x = -\frac{1}{2} \) into the first circle's equation \( S_1 \):

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

Calculating this:

\[ \frac{1}{4} + y^2 - 1 + 3y + 1 = 0 \]

This simplifies to:

\[ y^2 + 3y + \frac{1}{4} = 0 \]

### Step 3: Solve the quadratic equation for \( y \)

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

Here, \( a = 1, b = 3, c = \frac{1}{4} \):

\[ y = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot \frac{1}{4}}}{2 \cdot 1} = \frac{-3 \pm \sqrt{9 - 1}}{2} = \frac{-3 \pm \sqrt{8}}{2} = \frac{-3 \pm 2\sqrt{2}}{2} \]

This gives us:

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

Thus, the points \( P \) and \( Q \) are:

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

### Step 4: Calculate the lengths \( C_1P \) and \( C_1Q \)

Using the distance formula, the distance \( C_1P \):

\[ C_1P = \sqrt{\left(-1 - \left(-\frac{1}{2}\right)\right)^2 + \left(-\frac{3}{2} - \left(-\frac{3}{2} + \sqrt{2}\right)\right)^2} \]

Calculating this:

\[ = \sqrt{\left(-1 + \frac{1}{2}\right)^2 + \left(-\frac{3}{2} + \frac{3}{2} - \sqrt{2}\right)^2} \]
\[ = \sqrt{\left(-\frac{1}{2}\right)^2 + (-\sqrt{2})^2} \]
\[ = \sqrt{\frac{1}{4} + 2} = \sqrt{\frac{1}{4} + \frac{8}{4}} = \sqrt{\frac{9}{4}} = \frac{3}{2} \]

Since \( C_1P = C_1Q \), we have:

\[ C_1Q = \frac{3}{2} \]

### Step 5: Calculate the length of \( PQ \)

The distance \( PQ \):

\[ PQ = \sqrt{\left(-\frac{1}{2} - \left(-\frac{1}{2}\right)\right)^2 + \left(-\frac{3}{2} + \sqrt{2} - \left(-\frac{3}{2} - \sqrt{2}\right)\right)^2} \]
\[ = \sqrt{0 + (2\sqrt{2})^2} = \sqrt{8} = 2\sqrt{2} \]

### Step 6: Calculate the perimeter of triangle \( C_1PQ \)

The perimeter \( P \) is given by:

\[ P = C_1P + C_1Q + PQ = \frac{3}{2} + \frac{3}{2} + 2\sqrt{2} = 3 + 2\sqrt{2} \]

### Final Answer

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

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