Question
FInd the equation of circle which
passes through the points A(0,1), B(3,-2) and has its
centre
lying on
the line
y = x - 2
Solution
Mid-point of AB = [ (0+3) /2 , (1 + ( -2)) ] = (1.5, - 0.5)
Gradient of AB = ( -2 - 1) / (3 - 0) = - 1
Gradient of Perpendicular bisector of AB = - 1 / - 1 = 1
Subsituting midpoint (1.5, - 0.5 ) and gradient of 1 into
y = mx + c
- 0.5 = 1 (1.5) +c
c = - 2
So, the equation of the perpendicular bisector of AB
is
y = mx + c
y = 1x - 2
Hence, the equation of the
perpendicular bisector of AB is the SAME as the
line
y = x - 2 where the centre of the
circle lies.
THUS, the equation of the perpendicular bisector
of AB is the diameter of the circle
and the midpoint (1.5, - 0.5) is also the centre of the
circle.
Using A(0,1) and midpoint (1.5, - 0.5),
radius = square root [(1.5 - 0)^2 + (- 0.5 - 1)^2]
radius = square root (4.5)
Hence, the equation of the circle is (x - 1.5)^2
+ (y + 0.5)^2 = 4.5
