Respuesta :
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:
[tex]a(3^{2} + (-2)^2) + 3b - 2c + d = 0[/tex]
[tex]13a+ 3b - 2c + d = 0[/tex]
For (1, -2)
[tex]a(1^{2} + (-2)^2) + b - 2c + d = 0[/tex]
[tex]5a + b - 2c + d = 0[/tex]
For (0,0)
[tex]a(0^{2} + (0)^2) + 0b - 0c + d = 0[/tex]
[tex]d = 0[/tex]
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
