A circle has a center that falls on the line #y = 1/4x +4 # and passes through # ( 3 ,7 )# and #(7 ,1 )#. What is the equation of the circle?

1 Answer
Mar 3, 2018

#x^2+y^2-16x-12y+74=0# is the required equation of the circle.

Explanation:

Let #P(h,k)# be the center the circle.

Then, #k=h/4+4=(h+16)/4#....[1]

Also it passes through #(3,7)# and #(7,1)#

The equation of circle is given by

#rarr(x-h)^2+(y-k)^2=r^2# where #(x,y)# is passing point and #(h,k)# is center and #r# is the radius.

So, #rarr(3-h)^2+(7-k)^2=r^2#

#rarr9-6h+h^2+49-14k+k^2=r^2#

#rarrh^2+k^2-6h-14h+58=r^2#.....[2]

and #(7-h)^2+(1-k)^2=r^2#

#rarr49-14h+h^2+1-2k+k^2=r^2#

#rarrh^2+k^2-14h-2k+50=r^2#.....[3]

From [2 and [3], we have,

#rarrcancel(h^2+k^2)-6h-14h+58=cancel(h^2+k^2)-14h-2k+50#

#rarr8h-12k+8=0#

#rarr4*(2h-3k+2)=0#

#rarr2h-3k+2=0#....[4]

Putting the value of #k# from euation [1] to equation [4],

#rarr2h-3*(h+16)/4+2=0#

#rarr8h-3h-48+8=0#

#rarrh=8#

We have, #k=(h+16)/4#

#rarrk=(16+8)/4=6#

Now, #r^2=(7-h)^2+(1-k)^2=(7-8)^2+(1-6)^2=26#

The required equation is

#rarr(x-h)^2+(y-k)^2=r^2#

#rarr(x-8)^2+(y-6)^2=26#

#rarrx^2-16x+64+y^2-12y+36=26#

#rarrx^2+y^2-16x-12y+74=0#