A circle has a center that falls on the line #y = 7/9x +7 # and passes through # ( 4 ,5 )# and #(8 ,1 )#. What is the equation of the circle?

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Answer 1 · 2017-03-17T07:06:10

The equation of the circle is #(x-45)^2+(y-42)^2=3050#

The mid point of #(4,5)# and #(8,1)# is #(6,3)#

The slope of the line through #(4,5)# and #(8,1)# is

#m_1=(1-5)/(8-4)=-4/4=-1#

The slope of the line perpendicular to the line through #(4,5)# and #(8,1)# is

#m_2=1#

The equation of the line through #(6,3)# with a slope of #1# is

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

#y=x-6+3=x-3#

The center of the circle lies on the intersection of lines

#y=7/9x+7#

and

#y=x-3#

Therefore,

#7/9x+7=x-3#

#x(1-7/9)=7+3=10#

#2/9x=10#

#x=10*9/2=45#

and

#y=45-3=42#

So,

The center of the circle is #C=(45,42)#

The radius is

#r=sqrt((45-4)^2+(42-5)^2)#

#=sqrt(41^2+37^2)=sqrt3050#

#r^2=3050#

The equation of the circle is

#(x-45)^2+(y-42)^2=3050#

graph{((x-4)^2+(y-5)^2-1/16)((x-8)^2+(y-1)^2-1/16)(y-x+3)(y-7/9x-7)(x+y-9)=0 [-8.75, 21.25, -4.1, 10.9]}

graph{((x-45)^2+(y-42)^2-3050)(y-x+3)(y-7/9x-7)=0 [-90, 150, -20, 100]}