A circle has a center that falls on the line #y = 5/4x +5 # and passes through # ( 4 ,7 )# and #(2 ,5 )#. What is the equation of the circle?

The centre of the circle must lie on the perpendicular bisector of the line segment joining #(4, 7)# and #(2, 5)#.

Their midpoint is:

#((4+2)/2, (7+5)/2) = (3, 6)#

and the slope of the line segment joining them is:

#(5-7)/(2-4) = (-2)/(-2) = 1#

Hence the slope of the perpendicular bisector is:

#-1/color(blue)(1) = -1#

Hence the equation of the perpendicular bisector can be written:

#y - 6 = -1(x - 3)#

which simplifies to:

#y = 9-x#

This will intersect the given line when:

#9-x = y = 5/4x+5#

Multiply both ends by #4# to get:

#36-4x=5x+20#

Add #4x-20# to both sides to get:

#16 = 9x#

Hence:

#x = 16/9#

and:

#y = 9-x = 9-16/9 = 65/9#

So the centre of the circle is at: #(16/9, 65/9)#

So its equation can be written in the form:

#(x-16/9)^2 + (y-65/9)^2 = r^2#

where the radius #r# is yet to be determined.

Since this passes through #(4, 7)#, those values of #x# and #y# must satisfy the equation.

So we find:

#r^2 = (color(blue)(4)-16/9)^2 + (color(blue)(7)-65/9)^2#

#color(white)(r^2) = (20/9)^2 + (-2/9)^2#

#color(white)(r^2) = 404/81#

So the equation of the circle may be written:

#(x-16/9)^2 + (y-65/9)^2 = 404/81#

graph{(y-x-3)((x-4)^2+(y-7)^2-0.006)((x-2)^2+(y-5)^2-0.006)((x-16/9)^2 + (y-65/9)^2 - 404/81)((x-16/9)^2 + (y-65/9)^2 - 0.006)(y+x-9)(y-5/4x-5)=0 [-3.63, 6.37, 4.6, 9.6]}