A circle has a center that falls on the line #y = 5/8x +6 # and passes through # ( 5 ,2 )# and #(3 ,4 )#. What is the equation of the circle?

2 Answers
Dec 12, 2016

The equation is #(x-56/3)^2+(y-53/3)^2=20.8^2#

Explanation:

Let the center of the circle be #(a,b)#

Then the equation of the circle is

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

The points on the circle are #(5,2)# and #(3.4)#

Therefore,

#(5-a)^2+(2-b)^2=r^2#

and #(3-a)^2+(4-b)^2=r^2#

So,

#(5-a)^2+(2-b)^2=(3-a)^2+(4-b)^2#

#25-10a+a^2+4-4b+b^2=9-6a+a^2+16-8b+b^2#

#4-10a-4b=-6a-8b#

#4a-4b=4#

#a-b=1# this is the first equation

#(a,b)# lies on the line #y=5/8x+6#

So, #b=5/8a+6#

So, #8b=5a+48# this is the second equation

Solving for #(a,b)# with the 2 simultaneous equations

#8b=5(b+1)+48#

#3b=53#, #=>#, #b=53/3#

#a=b+1=53/3+1=56/3#

The center of the circle is #(56/3,53/3)#

The radius is #r^2=(5-a)^2+(2-b)^2#

#r^2=(5-56/3)^2+(2-53/3)^2#

#r=sqrt((41/3)^2)+(47/3)^2)#

#=20.8#

The equation of the circle is

#(x-56/3)^2+(y-53/3)^2=20.8^2#

graph{((x-56/3)^2+(y-53/3)^2-20.8^2)(y-5/8x-6)=0 [-105.4, 105.5, -52.8, 52.7]}

Dec 12, 2016

The equation is #(x-56/3)^2+(y-53/3)^2=20.8^2#

Explanation:

Let the center of the circle be #(a,b)#

Then the equation of the circle is

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

The points on the circle are #(5,2)# and #(3.4)#

Therefore,

#(5-a)^2+(2-b)^2=r^2#

and #(3-a)^2+(4-b)^2=r^2#

So,

#(5-a)^2+(2-b)^2=(3-a)^2+(4-b)^2#

#25-10a+a^2+4-4b+b^2=9-6a+a^2+16-8b+b^2#

#4-10a-4b=-6a-8b#

#4a-4b=4#

#a-b=1# this is the first equation

#(a,b)# lies on the line #y=5/8x+6#

So, #b=5/8a+6#

So, #8b=5a+48# this is the second equation

Solving for #(a,b)# with the 2 simultaneous equations

#8b=5(b+1)+48#

#3b=53#, #=>#, #b=53/3#

#a=b+1=53/3+1=56/3#

The center of the circle is #(56/3,53/3)#

The radius is #r^2=(5-a)^2+(2-b)^2#

#r^2=(5-56/3)^2+(2-53/3)^2#

#r=sqrt((41/3)^2)+(47/3)^2)#

#=20.8#

The equation of the circle is

#(x-56/3)^2+(y-53/3)^2=20.8^2#

graph{((x-56/3)^2+(y-53/3)^2-20.8^2)(y-5/8x-6)=0 [-105.4, 105.5, -52.8, 52.7]}