A circle has a center that falls on the line #y = 7/8x +1 # and passes through # ( 5 ,2 )# and #(3 ,6 )#. What is the equation of the circle?

2 Answers
Jul 2, 2017

The equation of the circle is #(x-8/3)^2+(y-10/3)^2=65/9#

Explanation:

Let #C# be the mid point of #A=(5,2)# and #B=(3,6)#

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

The slope of #AB# is #=(6-2)/(3-5)=(4)/(-2)=-2#

The slope of the line perpendicular to #AB# is #=1/2#

The equation of the line passing trrough #C# and perpendicular to #AB# is

#y-4=1/2(x-4)#

#y=1/2x-2+4=1/2x+2#

The intersection of this line with the line #y=7/8x+1# gives the center of the circle.

#1/2x+2=7/8x+1#

#7/8x-1/2x=2-1#

#3/8x=1#

#x=8/3#

#y=1/2*(8/3)+2=10/3#

The center of the circle is #(8/3,10/3)#

The radius of the circle is

#r^2=(5-8/3)^2+(2-10/3)^2#

#=(7/3)^2+(-4/3)^2#

#=65/9#

The equation of the circle is

#(x-8/3)^2+(y-10/3)^2=65/9#

graph{((x-8/3)^2+(y-10/3)^2-65/9)(y-7/8x-1)(y-1/2x-2)=0 [-9, 11, -1.96, 8.04]}

Jul 2, 2017

#3x^2+3y^2-16x-20y+33=0#

Explanation:

Let the equation of circle be #x^2+y^2+2gx+2fy+c=0#, whose center is #(-g,-f)# and radius is #sqrt(g^2+f^2-c)#.

As the centre is on #y=7/8x+1#, we have #-f=-7/8g+1#

or #7g-8f=8# ......................(1)

It also passes through #(5,2)# and #(3,6)#, we have

#5^2+2^2+10g+4f+c=0# or #10g+4f+c+29=0# ......................(2)

#3^2+6^2+6g+12f+c=0# or #6g+12f+c+45=0# ......................(3)

Subtracting (2) from (3), we get

#-4g+8f+16=0# ......................(4)

Adding (1) and (4), we have #3g=-8# or #g=-8/3#

and puuting this in (1), we get #-8f=8-7(-8/3)=8+56/3=80/3#

Hence #f=-10/3# and then putting #g# and #f# in (2), we get

#10(-8/3)+4(-10/3)+c+29=0# or #-80/3-40/3+c+29=0#

i.e. #c=-29+120/3=11#

Hence, equaton of circle is

#x^2+y^2-16/3x-20/3y+11=0#

or #3x^2+3y^2-16x-20y+33=0#

graph{(3x^2+3y^2-16x-20y+33)((x-5)^2+(y-2)^2-0.01)((x-3)^2+(y-6)^2-0.01)(y-(7x)/8-1)=0 [-7.83, 12.17, -1.94, 8.06]}