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.

1 Answer
Sep 14, 2017

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.