There are two types of equations of circles. They are
1. Standard equation of a circle
2. General equation of a circle
We can write an equation of a circle in a coordinate plane if we know its radius and the coordinates of its center. Suppose the radius of a circle is r and the center is (h,k).
Let (x,y) be any point on the circle. The distance between (x,y) and (h,k) is r.
So, we can use the Distance Formula.
Square both sides to find the standard equation of a circle with radius r and center (h,k).
If the center is the origin, then (h, k) = (0, 0)
So, the standard equation is
x^{2} + y^{2} = r^{2}
The equation given below is the general form of equation of a circle.
x^{2} + y^{2} + 2gx + 2fy + c = 0
Center = (-g, -f)
Radius = √[g^{2} + f^{2} - c]
Example 1 :
Write the standard equation of the circle whose center is (-4, 0) and radius is 7.
Solution :
Standard equation of a circle.
(x - h)^{2} + (y - k)^{2} = r^{2}
Plug (h, k) = (-4, 0) and r = 7.
[x - (-4)^{2}] + (y - 0)^{2} = 7^{2}
Simplify.
(x + 4)^{2} + y^{2} = 49
Example 2 :
Write the standard equation of the circle whose general equation is
x^{2} + y^{2 }- 4x + 6y - 12 = 0
Solution :
To find the standard equation of the circle, we need to know the center and radius.
Let us find the center and radius from the given general equation of the circle.
Comparing
x^{2} + y^{2} + 2gx + 2fy + c = 0
and
x^{2} + y^{2 }- 4x + 6y - 12 = 0,
we have
2g = - 4 -----> g = -2
2f = 6 -----> f = 3
c = -12
Center = (-g, -f) = (2, -3)
Radius = √[g^{2} + f^{2} - c]
Radius = √[(-2)^{2} + 3^{2} - (-12)]
Radius = √[4 + 9 + 12]
Radius = √25
Radius = 5
Standard equation of a circle :
(x - h)^{2} + (y - k)^{2} = r^{2}
Plug (h, k) = (2, -3) and r = 5.
(x - 2)^{2} + [y - (-3)^{2}] = 5^{2}
Simplify.
(x - 2)^{2} + (y + 3)^{2} = 25
Example 3 :
The point (1 ,2) is on a circle whose center is (5, -1). Write the standard equation of the circle.
Solution :
To find the standard equation of the circle, we need to know the center and radius. The center is already given and we need to find the radius.
Using distance formula, we have
Radius = √[(5 - 1)^{2} + (-1 - 2)^{2}]
Radius = √[4^{2} + (-3)^{2}]
Radius = √[16 + 9]
Radius = √25
Radius = 5
Standard equation of a circle :
(x - h)^{2} + (y - k)^{2} = r^{2}
Plug (h, k) = (5, -1) and r = 5.
(x - 5)^{2} + [y - (-1)^{2}] = 5^{2}
Simplify.
(x - 5)^{2} + (y + 1)^{2} = 25
If we know the equation of a circle, we can graph the circle by identifying its center and radius.
Example 4 :
The equation of a circle is
(x + 2)^{2} + (y - 3)^{2} = 9
Graph the circle.
Solution :
To graph a circle, we need to know the radius and center of the circle.
Rewrite the equation to find the center and radius.
(x + 2)^{2} + (y - 3)^{2} = 9
[x - (-2)^{2}] + (y - 3)^{2} = 3^{2}
So, the center is (-2, 3) and the radius is 3.
To graph the circle, place the point of a compass at (-2, 3), set the radius at 3 units and swing the compass to draw a full circle.
Example 5 :
A bank of lights is arranged over a stage. Each light illuminates a circular area on the stage. A coordinate plane is used to arrange the lights, using the corner of stage as the origin. The equation (x - 13)^{2} + (y - 4)^{2} = 16 represents one of the disks of light.
(i) Graph the disk of light.
(ii) Three actors are located as follows :
Henry is at (11, 4)
Jolene is at (8, 5)
Martin is at (15, 5)
Which actors are in the disk of light ?
Solution :
Part (i) :
To graph the disk of light, we need to know the center and radius of the circle.
Rewrite the equation to find the center and radius.
(x - 13)^{2} + (y - 4)^{2} = 16
(x - 13)^{2} + (y - 4)^{2} = 4^{2}
So, the center is (13, 4) and the radius is 4.
The circle with center at (13, 4) and radius 4 units shown below (Graph of the disk of light).
Part (ii) :
The graph given above shows that Henry and Martin are both in the disk of light.
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