How do you find an equation of the circle that passes through the pts (0,0) and (1,1) and has its center on the line y=2?

1 Answer
Jan 20, 2016

Answer:

#(x+1)^2+(y-2)^2=5#

Explanation:

If the center of the circle is on #y=2#
then the center of the circle is at #(x_c,2)# for some value #x_c#

The radius of the circle, #r#, is

  • The distance from #(x_c,2)# to #(0,0)#
    #color(white)("XXX")=sqrt((x_c-0)^2+(2-0)^2)=sqrt(x_c^2+4)#
  • The distance from #(x_c,2)# to #(1,1)#
    #color(white)("XXX")=sqrt((x_c-1)^2+(2-1)^2) = sqrt(x_c^2-2x_c+1+1)#

Therefore
#color(white)("XXX")x_c^2+4 = x_c^2-2x_c+2#

#color(white)("XXX")-2x_c= -2#

#color(white)("XXX")x_c= -1#

and since #r=sqrt(x_c^2+4)#
#color(white)("XXX")r=sqrt(5)#

The general equation of a circle with center #(a,b)# and radius #r# is
#color(white)("XXX")(x-a)^2+(y-b)^2=r^2#

By substituting #(-1,2)# for #(a,b)# and #sqrt(5)# for #r#, we get
#color(white)("XXX")(x+1)^2+(y-2)^2=5#
graph{(x+1)^2+(y-2)^2=5 [-5.8, 5.3, -0.764, 4.783]}