(a) Centre (h, k) and radius r
The equation is (x – h) 2 + (y - k) 2 = r 2 (1)
(b) General form of the equation of a circle is
x 2 + y 2 + 2gx + 2fy + c = 0 (2)
If we compare equation (1) and (2) we get the centre of (2) as (-g, -f) and radius =
(c) Equation of a circle on the segment joining (x 1 , y 1 ) and (x 2 , y 2 ) as the ends of a diameter is given by (x – x 1 )(x – x 2 ) + (y – y 1 )(y – y 2 ) = 0 (3)
(d) General equation of a circle which passes through A(x 1 , y1) and B(x 2 , y 2 ) may be written as
(x – x
1
)(x – x
2
) + (y – y
1
)( y – y2) + λ
(4)
(e) The equation of the circle through 3 non-collinear points A(x1, y1), B(x2, y2) and C(x3, y3) is
= 0
The point P(x 1 , y 1 ) lies outside or inside the circle
S = x 2 + y 2 + 2gx + 2fy + c = 0 according as
S
1
=
> or < 0