Find the centre of a circle passing through the points (6, -6),( 3,-7) and (6,3).
Answers
Let the centre of the circle be O and let the points be A(6, -6), B(3, -7), C(6, 3).
Then, OA=OB=OC (Radii of the same circle)
Let the point O be (x, y)
distance formula
= root(x2 - x1)^2 + (y2 - y1)^2
OA^2 = OB^2
=> (x - 6)^2 + (y + 6)^2 = (x - 3)^2 + (y + 7)^2
=> x^2 - 12x + 36 + y^2 + 12y + 36 = x^2 -6x + 9 + y^2 + 14y + 49
=> 72 - 12x + 12y = 14y - 6x + 58
=> 6x + 2y = 14
=> 3x + y = 7 -------------- ( 1 )
Similarly,
OB^2 = OC^2
(x - 3)^2 + (y + 7)^2 = (x - 6)^2 + (y - 3)^2
x^2 - 6x + 9 + y^2 + 14y + 49 = x^2 - 12x + 36 + y^2 - 6y + 9
-6x + 14y + 58 = -12x -6y + 45
-6x -20y = 13
6x + 20y = -13 --------- ( 2 )
Solve ( 1 ) and ( 2 )
3x + y = 7 ×2
6x + 20y = -13
6x + 2y = 14
6x + 20y = -13
- - +
-------------------------------
0 - 18y = 27
-------------------------------
y = -27/18
= -3/2
Put in eqn ( 1 )
3x - 3/2 = 7
3x = 7 + 3/2
3x = 15/2
x = 5/2
Thus the centre of the circle is O ( -3/2, 5/2)