A circle has a center that falls on the line #y = 2x +7 # and passes through #(4 ,4 )# and #(1 ,2 )#. What is the equation of the circle?

The standard Cartesian form for the equation of a circle is:

#(x - h)^2 + (y-k)^2=r^2" [1]"#

where #(x,y)# is any point on the circle, #(h,k)# is the center-point and #r# is the radius.

Use equation [1] and the two points to write two equations:

#(4 - h)^2 + (4-k)^2=r^2" [2]"#
#(1 - h)^2 + (2-k)^2=r^2" [3]"#

The third equation is written by substituting the center-point to the given line:

#k = 2h+7" [4]"#

Expand equations [2] and [3]:

#16 - 8h+ h^2 + 16- 8k+k^2=r^2" [5]"#
#1 - 2h+h^2 + 4-4k+k^2=r^2" [6]"#

Subtract equation [6] from equation [5]:

#27-6h-4k=0" [7]"#

Use equation [4] to substitute into equation [7]:

#27-6h-4(2h+7)=0#

#27-6h-8h-28=0#

#-14h-1 = 0#

#h = -1/14#

Use equation [4] to find the value of k:

#k = 2(-1/14)+7#

#k = -2/14+98/14#

#k = 96/14#

Use equation [3] to find the value of r:

#(1 - -1/14)^2 + (2-96/14)^2=r^2#

#(15/14)^2+ (-68/14)^2 = r^2#

#r = sqrt(4849)/14#

Use equation [1] to write the equation of the circle:

#(x+1/14)^2 + (y-96/14)^2=(sqrt(4849)/14)^2#
graph{(27-6x-4y)(2x-3y+4)(2x-y+7)((x-4)^2+(y-4)^2-0.05) ((x-1)^2+(y-2)^2-0.05) ((x+1/14)^2+(y-96/14)^2-4849/196)=0 [-10.09, 9.91, 1.2, 11.2]}