How do you write the standard form of the equation of a circle that passes through the points at (0,8) (8,0) AND (16,8)?

1 Answer
May 16, 2018

Answer:

#x^2+y^2-16x-16y+64=0#

Explanation:

Let the equation of circle be #x^2+y^2+2gx+2fy+c=0#

as it passes through #(0,8)#, #8,0)# and #(8,16)#, these points satisfy the above equation and we should have

#0^2+8^2+0*g+16f+c=0# i.e. #16f+c=-64# (A)

#8^2+0^2+16g+0*f+c=0# i.e. #16g+c=-64# (B)

and #16^2+8^2+32g+16f+c=0# i.e. #32g+16f+c=-320# (C)

Subtracting (A) and (B) from (C) we get

#32g=-256# or #g=-8#

and #16g+16f=-256# or #g+f=-16# and as #g=-8#, #f=-8#

and hence putting these value in (A), we get

#-128+c=-64# or #c=64#

Hence equation of circle is #x^2+y^2-16x-16y+64=0#

graph{(x^2+y^2-16x-16y+64)(y^2+(x-8)^2-0.07)(x^2+(y-8)^2-0.07)((x-16)^2+(y-8)^2-0.07)=0 [-12.83, 27.17, -3.12, 16.88]}