A circle has a center that falls on the line #y = 3/8x +5 # and passes through # ( 7 ,3 )# and #(2 ,1 )#. What is the equation of the circle?

There are a few techniques that can be used: Here I will use just algebra:

The general equation of a circle of centre #(a,b)# and radius #r# is:

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

The centre #(a,b)#lies on the given line #y=3/8x+5#; thus

# b=3/8a+5 # ..... [A]

The circle passes through #(7,3)#

# (7-a)^2+(3-b)^2=r^2#
# :. 49-14a+a^2+9-6b+b^2=r^2 #
# :. 58-14a+a^2+6b+b^2=r^2 # ..... [B]

The circle passes through #(2,1)#

# (2-a)^2+(1-b)^2=r^2#
# :. 4-4a+a^2 +1 -2b +b^2 =r^2#
# :. 5-4a+a^2 -2b +b^2 =r^2# ..... [C]

Solving for #a# and #b#

Eq [B] - Eq[C]:

# 53 -10a-4b = 0 #

And using [A] we get:

# 4b=3/2a+20 #

And so:

# 53 -10a-3/2a-20 = 0 #
# :. 33 -23a/2 = 0 #
# :. a=66/23 ~~ 2.86#

Subs #a=66/23# into Eq [A]:

# b=3/8*66/23+5 #
# :. b=559/92 ~~ 6.08 #

Solving for #r#

Substitute #a=66/23# and #b=559/92 # into Eq [B] we get:

# r^2 = 5-4(66/23)+(66/23)^2 -2(559/92) +(559/92)^2 #
# \ \ \ = 5-264/23+4356/529 -559/46 +312481/8464 #
# \ \ \ = 224489/8464 => r ~~ 5//15#

So the equation is:

# (x-66/23)^2+(y-559/92)^2=224489/8464 #

And we can verify the solution graphically: