Find the equation of the circle that passes through the points...

Question

Find the equation of the circle that passes through the points (7, 3) and (11, -1) and has its centre lying on the line 2x+y=7.

Answer 1

circle through (7,3) and (11,-1)

general formula is (x-h)^2 + (y-k)^2 = r^2

(7-h)^2 + (3-k)^2 = r^2 is Equation I

(-1-h)^2 + (11-k)^2 = r^2 is Equation II

center lies on 2x+y = 7

the point (h,k) is on the line 2x+y=7

2h+k=7 is Equation III

that's 3 equations 3 unknowns

solve with elimination and substitution or matrix algebra

49-14h^2 +h^2 + 9-6k+k^2 = r^2

1+2h +h^2 + 121 -22k +k^2 = r^2

subtract

48-16h -112 +16k = 0

16h -16k = -64

h-k = -4

h=k-4

2h+k=7

2(k-4)+k = 7

2k -8 +k = 7

3k = 15

k = 15/3 = 5

h= k-4 = 5-4 = 1

center is (1,5)

(7-1)^2 + (3-5)^2 = r^2

36 +4 = r^2

radius = sqr40 = 2sqr10

the circle's equation is

(x-1)^2 + (y-5)^2 = 40