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

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Answer 1 · 2016-06-28T05:41:09

#2x^2 + 2y^2 + 100x + 159y - 1643 = 0#

General equation of circle can be represented as #x^2+y^2-2hx-2ky+c=0# -> #Equation 1#

where #(h,k)# represents the center of the circle , #r #- radius of the circle, #c = h^2 + k^2 - r^2#

As we know the given points # (4,7) and (7,5)# lie on the circle, they must satisfy Equation 1.

Putting the given points # (4,7) and (7,5)# in the above equation of circle (1) , we get :

#4^2+7^2-2h*4-2k*7+c=0#

=># -8h - 14k + c = -65# =>

# 8h + 14k - c = 65# -> # Equation 2#

#7^2+5^2-2h*7-2k*5+c=0#

=>#-14h - 10k +c=-74# =>

# 14h + 10k - c=74# -> #Equation 3#

Since the centre of the circle is #(h,k)# lies on the line #y = 7/4x + 4#, #(h,k)# must satisfy the equation of line.

#k = 7/4*(h) + 4# =>

#7h - 4k = -16# -> #Equation 4#

Subtract Equation 2 and 3

=> #(14h - 8h) + (10k - 14k) + (-c+c) = # 74 - 65#

#6h -4k = 9# -> #Equation 5#

Add Equation 4 and 5

=>#(-7h+6h) + (4k -4k) = (16+9)# -> #Equation 5#

# -h = 25 => h= -25#

#-4k= 9 - (6*(-25)#

#k = -159/4#

Solving for c in Equation 3, we get #c = 14*(-25) + 10*(-159/4) - 74#

#c= -3286/4# = #-1643/2#

Circle equation is:

#x^2+y^2-2(-25)x-2(-159/4)y- 1643/2=0#

#x^2+y^2+50x-(-159/2)y - 1643/2=0#

#2x^2 + 2y^2 + 100x + 159y - 1643 = 0#