Write a systems of equations, letting #(x, y)# be the centre. The logic is the following: since the two points given lie on the circumference of the circle, the distance to the centre is equal with both points, so let equation 1 be #sqrt((x - 5)^2 + (y - 2)^2) = sqrt((x - 3)^2 + (y - 2)^2)#
The centre lies on the line #y = 2/3x + 7#, so it makes sense that our second equation is #y = 2/3x + 7#.
Let's now solve using substitution.
#sqrt((x - 5)^2 + (2/3x + 7 - 2)^2 ) = sqrt((x - 3)^2 + (2/3x + 7 - 2)^2)#
#x^2 - 10x + 25 + 4/9x^2 + 20/3x + 25 = x^2 - 6x + 9 + 4/9x^2 + 20/3x + 25#
#-4x = -16#
#x = 4#
#y = 2/3(4) + 7 = 8/3 + 7 = 29/3#
Hence, the centre has coordinates of #(4, 29/3)#.
Now, we need to determine the measure of the radius, again using the distance formula.
#r = sqrt((5 - 4)^2 + (29/3 - 2)^2)#
#r = sqrt(1 + 529/9)#
#r = sqrt(538/9)#
The most commonly used form of the equation of the circle is #(x - a)^2 + (y - b)^2 = r^2#, where #(a, b)# is the centre and #r# the radius.
Thus, the equation of the circle is #(x - 4)^2 + (y - 29/3)^2 = 538/9#
Hopefully this helps!
