A circle has a center that falls on the line #y = 5/8x +6 # and passes through # ( 1 ,4 )# and #(2 ,9 )#. What is the equation of the circle?

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

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

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

Use equation [1] and the points #(1, 4) and (2,9)# to write two equations:

#(1 - h)^2 + (4 - k)^2 = r^2" [2]"#
#(2 - h)^2 + (9 - k)^2 = r^2" [3]"#

Expand the squares on the left sides of equations [2] and [3], using the pattern, #(a - b)^2 = a^2 - 2ab + b^2#:

#1 - 2h + h^2 + 16 - 8k + k^2 = r^2" [4]"#
#4 - 4h + h^2 + 81 - 18k + k^2 = r^2" [5]"#

Subtract equation [5] from equation [4]:

#1 - 2h + h^2 + 16 - 8k + k^2 - (4 - 4h + h^2 + 81 - 18k + k^2) = r^2 - r^2#

#1 - 2h + h^2 + 16 - 8k + k^2 - 4 + 4h - h^2 - 81 + 18k - k^2 = 0#

#1 - 2h + 16 - 8k - 4 + 4h - 81 + 18k = 0#

#2h + 10k = 68#

#h + 5k = 34" [6]"#

Substitute h for x and k for y into the given equation, #y = 5/8x + 6#:

#k = 5/8h + 6" [7]"#

Substitute the right side of equation [7] for k into equation [6]:

#h + 5(5/8h + 6) = 34#

Solve for h:

#h + 25/8h + 30 = 34#

#33/8h = 4#

#h = 32/33#

Substitute this value for h into equation [7]:

#k = 5/8(32/33) + 6#

#k = 218/33#

Substitute the values for h and k into equation [3]:

#(2 - 32/33)^2 + (9 - 218/33)^2 = r^2#

Solve for r:

#(66/33 - 32/33)^2 + (297/33 - 218/33)^2 = r^2#

#(34/33)^2 + (79/33)^2 = r^2#

#r^2 = 7397/33^2#

#r = sqrt(7397)/33#

Substituting the values for, h, k, and r into equation [1], we obtain the equation:

#(x - 32/33)^2 + (y - 218/33)^2 = (sqrt(7397)/33)^2#

The following is a graph of the center point, the two given points, the given line and the circle: