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

Question

1315 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-30T18:11:11

#(x - 29/9)^2+(y-163/18)^2 = (5sqrt(941)/18)^2#

The standard Cartesian form for the equation of a circle is:

#(x - h)^2+(y-k)^2 = r^2" [1]"#

where, #(x,y)# is any point on the circle, #(h,k)# is the center point and #r# is the radius.

We may use equation [1] and the points, #(8,2)# and #(6,1)# to write 2 unique equations:

#(8 - h)^2+(2-k)^2 = r^2" [2]"#
#(6 - h)^2+(1-k)^2 = r^2" [3]"#

To obtain a third unique equation, we substitute the center point, #(h,k)#, into the given linear equation:

#k = 5/2h+1" [4]"#

Expand the squares within equations [2] and [3]:

#64 - 16h+h^2+4-4k+k^2 = r^2" [2.1]"#
#36 - 12h+h^2+1-2k+k^2 = r^2" [3.1]"#

Subtract equation [2.1] from equation [3.1]:

#4h+2k -31=0" [5]"#

Substitute equation [4] into equation [5]:

#4h+2(5/2h+1) -31=0#

#4h+5h+2-31=0#

#h= 29/9#

Substitute this into equation [4], to obtain the value of k:

#k = 5/2(29/9)+1#

#k = 163/18#

One may substitute the values of h and k into either equation [2] or [3] to obtain the value of r; I shall use equation [2]:

#(8 - 29/9)^2+(2-163/18)^2 = r^2#

#(144/18 - 58/18)^2+(36/18-163/18)^2 = r^2#

#(86/18)^2+(-127/18)^2 = r^2#

#r= sqrt(23525)/18#

#r = 5sqrt(941)/18#

Substitute the values of #h, k , and r# into equation [1], to obtain the equation of the circle:

#(x - 29/9)^2+(y-163/18)^2 = (5sqrt(941)/18)^2#