Mathematics, 22.10.2019 02:30 Northshoreranch

Write the general equation for the circle that passes through the points: (0,0) (6,0) (0,-8)

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Answer from: lobatospitones

(x-3)^2+(y+4)^2=25

Step-by-step explanation:

We start by writing the general form of the equation of a circle of radius R centered at (x_0,y_0) and creating three equations (one for each unknown: x_0 , y_0 , and the radius R):

(x-x_0)^2+(y-y_0)^2=R^2

1) If the circle passes through (0,0) then we should have that the equation above holds true:

$(x-x_0)^2+(y-y_0)^2=R^2\\(0-x_0)^2+(y-y_0)^2=R^2\\x_0^2+y_0^2=R^2\\R^2-x_0^2-y_0^2=0$

2) If the circle passes through (6,0) then we should have that the equation above holds true, and we also can use the important result from part 1) ( R^2-x_0^2-y_0^2=0 ) to solve for x_0 :

$(x-x_0)^2+(y-y_0)^2=R^2\\(6-x_0)^2+(y-y_0)^2=R^2\\36-12x_0+x_0^2+y_0^2=R^2\\-12x_0+36=R^2-x_0^2-y_0^2=0\\36=12x_0\\x_0=\frac{36}{12} =3$

3) If the circle passes through (0,-8) then we should have that the general equation of the circle above holds true, and we also can use the important result from part 1) ( R^2-x_0^2-y_0^2=0 ) to solve for y_0 :

$(x-x_0)^2+(y-y_0)^2=R^2\\(0-x_0)^2+(-8-y_0)^2=R^2\\x_0^2+64+16y_0+y_0^2=R^2\\64+16y_0=R^2-x_0^2-y_o^2=0\\16y_0=-64\\y_0=-\frac{64}{16}= -4$

4) and finally, we use the results for x_0 and y_0 of parts 2) and 3) back into the equation in part 1) to solve for R^2 :

$R^2-x_0^2-y_0^2=0\\R^2-(3)^2-(-4)^2=0\\R^2-9-16=0\\\R^2=9+16=25\\$

Then, replacing x_0 , y_0 , and R^2 for the values we found, the equation of the circle becomes:

$(x-x_0)^2+(y-y_0)^2=R^2\\(x-3)^2+(y+4)^2=25$

Answer from: Quest

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step-by-step explanation:

Answer from: Quest

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