The equation of the circle whose one diameter is PQ, where the ordinates of P, Q are the roots of the equation `x^(2)+2x-3=0` and the abscissae are the roots of the equation `y^(2)+4y-12=0` is

Options

A. `x^(2)+y^(2)+2x+4y-15=0`
B. `x^(2)+y^(2)-4x-2y-15=0`
C. `x^(2)+y^(2)+4x+2y-15=0`
D. none of these

Answer

To find the equation of the circle whose diameter is defined by the points P and Q, we need to follow these steps:

### Step 1: Find the roots of the equations for the ordinates and abscissae.
1. **Find the roots of the equation** \(x^2 + 2x - 3 = 0\):
- This can be factored as \((x + 3)(x - 1) = 0\).
- Thus, the roots are \(x_1 = -3\) and \(x_2 = 1\).

2. **Find the roots of the equation** \(y^2 + 4y - 12 = 0\):
- This can be factored as \((y + 6)(y - 2) = 0\).
- Thus, the roots are \(y_1 = -6\) and \(y_2 = 2\).

### Step 2: Identify the coordinates of points P and Q.
- The coordinates of points P and Q can be defined as:
- \(P(-3, -6)\)
- \(Q(1, 2)\)

### Step 3: Use the formula for the equation of a circle with diameter PQ.
The equation of a circle with diameter endpoints \(P(x_1, y_1)\) and \(Q(x_2, y_2)\) is given by:
\[ (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \]
Substituting the coordinates:
\[ (x + 3)(x - 1) + (y + 6)(y - 2) = 0 \]

### Step 4: Expand the equation.
1. Expand \((x + 3)(x - 1)\):
\[ x^2 - x + 3x - 3 = x^2 + 2x - 3 \]

2. Expand \((y + 6)(y - 2)\):
\[ y^2 - 2y + 6y - 12 = y^2 + 4y - 12 \]

### Step 5: Combine the expanded equations.
Combine the two expanded results:
\[ (x^2 + 2x - 3) + (y^2 + 4y - 12) = 0 \]
This simplifies to:
\[ x^2 + y^2 + 2x + 4y - 15 = 0 \]

### Final Answer:
The equation of the circle is:
\[ x^2 + y^2 + 2x + 4y - 15 = 0 \]

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