Question

The general equation for a circle is a(x^2 + y^2) + 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

Answer:

The equation for this circle is 2c(x²+y²) - 8cx + cy = 0

Step-by-step explanation:

We can solve this problem replacing each of the points in the general equation and then solving the system of equations.

The fact that the circle passes by the point (3,-2) means that when x = 3, y = -2. So:

$a(3^{2} + (-2)^2) + 3b - 2c + d = 0$

$13a+ 3b - 2c + d = 0$

For (1, -2)

$a(1^{2} + (-2)^2) + b - 2c + d = 0$

$5a + b - 2c + d = 0$

For (0,0)

$a(0^{2} + (0)^2) + 0b - 0c + d = 0$

$d = 0$

Now we have to find a,b,c, from the following equations

13a+ 3b - 2c = 0

5a + b - 2c = 0

Since we have variables and 2 equations, i am going to write a and b as functions of c.

13a + 3b = 2c

5a + b = 2c

I will write b in the second equation as a function of a and c, and replace in the first equation

b = 2c - 5a

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13a + 3(2c-5a) = 2c

13a + 6cc -15a = 2c

-2a = -4c *(-1)

2a = 4c

a = 2c

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b = 2c - 5a

b = 2c - 5(2c)

b = -8c

The solution for the system is (a,b,c) = (2c, -8c, c). So the equation for this circle is 2c(x²+y²) - 8cx + cy = 0