Question

The general equation for a circle is a(x² + y²) + bx + cy + d = 0. There is exactly one circle passing through the points (3,-2),(1, -2), and (0,0). Find an equation for this circle. (x2 + y2) + x+ y + = 0

Answer 1

The equation of the circle that passes through the points (3,-2), (1, -2), and (0,0) is found by solving a system of equations derived from the general circle equation, resulting in the equation (x - 2)² + (y + 2)² = 2.

Step-by-step explanation:

To find the equation of the circle that passes through the points (3,-2), (1, -2), and (0,0), we need to solve a system of equations to find the values of a, b, and c in the general equation of a circle, which is a(x² + y²) + bx + cy + d = 0. By plugging in the points into the equation:

• For (3, -2): 9a + 3b - 2c + d = 0
• For (1, -2): a + b - 2c + d = 0
• For (0, 0): d = 0

Since we already know that d is 0 from the third equation, we can substitute into the first two and solve the system of linear equations.

After solving the equations, we end up with a circle equation (x - 2)² + (y + 2)² = 2, which represents the circle passing through the given points.