Given : The center of the a circle lies on the equation of line y=3/8x+8 and passes through points A(a_x, a_y) = A(7,4) and B(b_x, b_y)=B(2,9)
Required : Equation of the circle
Solution strategy: a) Equation of the circle centered at O(x_c, y_c) => (x+x_c)^2 + (y+y_c)^2= r^2
The center of the circle (x_c, y_c)=(x, 3/8x+8)
b) From the distance formula:
r^2 = (x-x_c)^2 + (y-y_c)^2 such that
A=>x: x= a_x=7 and y: y=a_y=4
B=>x: x= b_x=2 and y: y=b_y=9
We start with b) and writing two Distance Formula equations for A and B respectively:
A=> r^2= (x-7)^2+(3/8x+8-4)^2
=(x-7)^2+(3/8x+4)^2
B=> r^2= (x-2)^2+(3/8x+8-9)^2
= (x-2)^2+(3/8x-1)^2
Now both A and B lie on the circle so thy are equal to one another, i.e r^2=r^2 thus,
(x-7)^2+(3/8x+4)^2 = (x-2)^2+(3/8x-1)^2 Rearrange and clean up 64(x-7)^2 + (3x+32)^2 = 64(x-2)^2 + (3x-8)^2
Expand and eliminate the square in x terms:
64(x^2-14x+49) + (9x^2+192x+1024) =
64(x^2-4x+4) + (9x^2-48x+64)
cancel(73x^2)-704x+4160=cancel(73x^2)-304x+320 Solve for x
x=48/5. Now insert x=48/5 on the equation of line y=3/8x+8
y=3/8*48/5 +8 = 58/5
Thus the center, O(x_c, y_c)= O(48/5, 58/5)
To verify calculate the radiuses, bar(OA) and bar(OB)
(bar(OA))^2=(x_c-7)^2+(y_c-4)^2=(48/5-7)^2+(58/5-4)^2
(bar(OA))^2 = 169/25 + 1444/25
(bar(OB))^2=(x_c-2)^2+(y_c-9)^2=(48/5-2)^2+(58/5-9)^2
(bar(OB))^2 = 1444/25 + 169/25
(bar(OA))^2=(bar(OB))^2 = r^2
Now from strategy a) we have
(x+x_c)^2+(y+y_c)^2= r^2 replace (x_c, y_c) and r^2 by (48/5, 58/5) and r^2=1613/25
(x+48/5)^2+(y+58/5)^2= 1613/25
