Question

Can someone help me out please? Thanks!

Find a quadratic model for the set of values: (-2, -20), (0, -4), (4, -20). Show your work.

Answer 1

See below.

Explanation:

Given

`p_1 = (x_1,y_1)=(-2,-20)`
`p_2 = (x_2,y_2)=(0,-4)`
`p_3 = (x_3,y_3)=(4,-20)`

We know that along three non aligned points passes an unique circle, which is a quadratic curve. The generic circle centered at `p_0` with radius `r` is

`(x-x_0)^2+(y-y_0)^2=r^2` or

`x^2+y^2-2x x_0 -2y y_0 + x_0^2+y_0^2= r^2`

Applying this equation at `p_1,p_2,p_3` we have

`{(x_1^2+y_1^2-2x_1 x_0 -2y_1 y_0 + x_0^2+y_0^2= r^2),(x_2^2+y_2^2-2x_2 x_0 -2y_2 y_0 + x_0^2+y_0^2= r^2),(x_3^2+y_3^2-2x_3 x_0 -2y_3 y_0 + x_0^2+y_0^2= r^2):}`

Subtracting the first from the second and the first from the third we obtain

`{(x_2^2+y_2^2-x_1^2-y_1^2+2x_0(x_1-x_2)+2y_0(y_1-y_2)=0),(x_3^2+y_3^2-x_1^2-y_1^2+2x_0(x_1-x_3)+2y_0(y_1-y_3)=0):}`

so we can solve for `x_0, y_0` obtaining the center coordinates.

Once we have `x_0,y_0` substituting those values into

`x_1^2+y_1^2-2x_1 x_0 -2y_1 y_0 + x_0^2+y_0^2= r^2` we obtain also the value for `r`

The calculation work is left as an exercise.