The circle
If you take a piece of string and secure one end, then take the other end and
stretch it out and move it about 360 degrees youâ€™ll have a circle. This means
all points on the edge of the circle are an equidistant from a fixed point (the
centre)
The other option is to think about the circle in terms of two stacked cones.
Take a horizontal cut, and it is the circle
The point P represents any point along the edge of the circle.
The length r, is equidistant from the centre to any point on the edge of the
circle.
The equation is for a circle is r^{ 2} = x^{ 2} + y^{ 2}.
This is the standard form of the equation, which is centered at the origin.
For circles not centered at the origin:
(x + h)^{ 2} + (y  k)^{ 2 }= r^{ 2}, where h & k are
the coordinates of the origin
Example 1
What is the equation of a circle (centred at the origin) with a radius of 5?
Since r = 5, we substitute that into the equation
x^{ 2} + y ^{2 }= (5)^{ 2}
x^{ 2} + y ^{2 }= 25
Example 2
Write an equation of a circle with a radius of 4, and a centre @ (3,1). State
the domain and range of the circle.
Using the equation (x + h)^{ 2} + (y  k)^{ 2 }= r^{ 2},
the radius 4, and the centre (3,1)
(x  3)^{ 2} + (y  (1))^{ 2 }= 4^{ 2}
(x  3)^{ 2} + (y +1)^{ 2 }= 16^{ 2}
Domain : { x  1 < x < 7; x
}
Range : { y  5 < y < 3; y
}
Example 3
What is the equation of the circle in standard form, with centre at (2,3)
and through the
point P(4,1)
Draw a picture to assist in the process.
Using the Pythagorean Theorem, we can find the length of the radius.
r^{ 2 }= (4  (2))^{ 2} + (1  3)^{ 2 }
r^{ 2 }= 36 + 4
r^{ 2 }= 40
Since we know the centre, the equation is (x + 2)^{ 2} + (y  3)^{ 2
}= 40
