A circle has a center that falls on the line #y = 2/9x +8 # and passes through # ( 2 ,5 )# and #(1 ,4 )#. What is the equation of the circle?

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Answer 1 · 2018-01-19T13:51:41

Equation of the circle is

#color(blue)((x - (28/11))^2 + (y - (38/11))^2 = 1.639)#

Given
#y = (2/9)x + 8# and the line is passing through the center.

Let (h, k) be the coordinates of the circle center.

#9h - 2k = 16# Eqn (1)

#(x-h)^2 + (y-k)^2 = r^2#
Where (h,k) coordinates of the center and r the radius.

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(2,5) & (1,4) are points on the circumference of the circle and hence

#(2-h)^2 + (5-k)^2 = r^2 = (1-h)^2 + (4-k)^2#

#4 - 4h + cancel(h^2) + 25 -10k + cancel(k^2) = 1 - 2h + cancel(h^2) + 16 - 8k + cancel(k^2)#

#4 + 25 - 1 - 16 = -2h + 4h - 8k + 10k#

#2h + 2k = 12#

#h + k = 6# Eqn (2)

Solving Eqns (1), (2) we get the coordinates of the center.

#O (28/11, 38/11)#

#r = sqrt((2-(28/11))^2 + (5-(38/11))^2) ~~ 1.639#

Equation of the circle is

#color(blue)((x - (28/11))^2 + (y - (38/11))^2 = 1.639)#