A circle has a center that falls on the line #y = 3/8x +5 # and passes through # ( 7 ,3 )# and #(2 ,9 )#. What is the equation of the circle?

The center should be equidistant from #(7,3)# and #(2,9)#. Use the Pythagoras Theorem to find the distance between two points.

#sqrt{(x-7)^2+(y-3)^2} = sqrt{(x-2)^2+(y-9)^2}#

#(x-7)^2 + (y-3)^2 = (x-2)^2 + (y-9)^2#

#(x^2-14x+49) + (y^2-6y+9) = (x^2-4x+4) + (y^2-18y+81)#

#-14x+49 - 6y+9 = -4x+4 -18y+81#

#12y = 10x+27#

The result is an equation of a straight line, that all the points on it are equidistant from #(7,3)# and #(2,9)#. We will find where it intersects with the line #y = 3/8 x + 5# to locate the center of the circle.

#y = 3/8 x + 5#
#12y = 10x+27#

Solve them simultaneously.

#12(3/8 x + 5) = 10x+27#

#9 x + 120 = 20x+54#

#66 = 11x#

#x = 6#

#y = 3/8 (6) + 5#

#= 29/4#

The center of the circle is at #(6,29/4)#. You can verify that it is indeed equidistant from #(7,3)# and #(2,9)#.

#sqrt{(6-7)^2+(29/4-3)^2} = sqrt305/4#

#sqrt{(6-2)^2+(29/4-9)^2} = sqrt305/4#

Bear in mind that the radius is the distance from the center to any point on the circumference. Therefore, the radius is #sqrt305/4#.

For a circle with center at #(a,b)# and radius #r#, the cartesian equation is

#(x-a)^2 + (y-b)^2 = r^2#

Our circle has center at #(6,29/4)# and radius #sqrt305/4#, so its equation is

#(x-6)^2 + (y-29/4)^2 = (sqrt305/4)^2#

#16(x-6)^2 + (4y-29)^2 = 305#