Pre-Calc help:
Write the general equation for the circle that passes through the points (1, 1), (1, 3), and (9, 2).
You must include the appropriate sign (+ or -) in your answer. Do not use spaces in your answer.
8x² + 8y²___x___y = 0


Respuesta :

Answer:

[tex]8x^2+8y^2-79x-32y+95=0[/tex]

Step-by-step explanation:

Let the equation be : [tex]x^2+y^2+2gx+2fy+c=0[/tex]

The point (1,1) must satisfy this equation.

[tex]1^2+1^2+2g*1+2f*1+c=0\\2g+2f+c=-2---(1)[/tex]

Similarly, the point (1,3) must satisfy:

[tex]1^2+3^2+2g*1+2f*3+c=0\\1+9+2g+6f+c=0\\2g+6f+c=-10---(2)[/tex]

Also for the point (9,2), we have:

[tex]9^2+2^2+2g*9+2f*2+c=0\\81+4+18g+4f+c=0\\18g+4f+c=-85---(3)[/tex]

Solving the equations (1), (2) and (3) simultaneously, we get:

[tex]g=-\frac{-79}{16},f=-2,c=\frac{95}{8}[/tex]

We Substitute this values to get:

[tex]x^2+y^2+2(\frac{-79}{16})x+2(-2)y+\frac{95}{8}=0[/tex]

[tex]x^2+y^2-(\frac{79}{8})x-4y+\frac{95}{8}=0[/tex]

Multiply through by 8 to get:

[tex]8x^2+8y^2-79x-32y+95=0[/tex]