Back to: Further Mathematics SS2
Definition:
A circle is defines as the locus of point equidistant from a fixed point. A circle is completely specified by the centre and the radius.
Equation of a circle with centre (a,b) and radius r.
From (x – a)2 + (y – b)2 = r2
r2 – 2ax + a2 + y2 – 2by + b2 – r2 = 0
x2 + y2 – 2ax – 2by + a2 + b2 – r2 = 0
The above equation can be written as x2 + y2 + 2gx + 2fy + c = 0
Where a = – g, b = -f, c = – a2 + b2 – r2
Hence: x2 + y2 + 2gx + 2fy + c = 0 is called the general equation of a circle. observe the following about the general equation
i. It is a second degree equation in x and y
iiThe co-efficient of x2 and y2 are equal
iiiIt has no xy term
Examples:
1.Find the equation of a circle of centre (3, -2) radius 4 unit
Solution:
a = 3, b = -2 and r = 2
(x-a)2 + (y-b)2 = r2
(x-3)2 + (y+2)2 = 42
x2 – 6x + 9 + y2 + 4y + 4 = 16
x2 + y2 – 6x + 4y + 9 + 4 – 16 = 0
x2 + y2 – 6x + 4y – 3 = 0
Find the centre and radius of a circle whose equation is x2 + y2 – 6x + 4y – 3 = 0
Solution:
x2 + y2 – 6x + 4y – 3 = 0
x2 – 6x + y2 + 4y = + 3
Complete the square for x and y
x2 – 6x + 9 + y2 4y + 4 = 3 + 9 + 4
(x – 3)2 + (y + 2)2 = 16
Compare with (x – a)2 + (y – b) = r2
Equation of a circle passing through 3 points
Find the equation of the circumcircle of the triangle whose vertices are A (2,3) B (5,4) and C (3,7)
Solution:
The equation of the circle x2 + y2 + 2gx + 2fy + c = 0
22 + 32 + 4g + 6f + c = 0
52 + 42 + 10g + 8f + c = 0
32 + 72 + 6f + 14f + c = 0
Simplify the 3 equations
f = – 107 / 22
g = – 67 / 22
c = – 312 / 11
Hence, the equation of the circle is
x2 + y2 + 2 x + 2 y + = 0
11x2 + 11y2 – 67x – 107y + 312 = 0
a = 3, b = -2, r2 = 16,
r = = 4
hence the centre is (3, -2) and the radius is 4 unit
Evaluation:
1.Find the equation of the circle (-1 -1) and radius 3
Find the centre and radius of the circle x2 + y2 – 6x + 14y + 49 = 0
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