The Circle In Standard Form

The Circle in Standard Form

A circleA circle is the set of points in a plane that lie a fixed distance from a given point, called the center. is the set of points in a plane that lie a fixed distance, called the radiusThe fixed distance from the center of a circle to any point on the circle., from any point, called the center. The diameterThe length of a line segment passing through the center of a circle whose endpoints are on the circle. is the length of a line segment passing through the center whose endpoints are on the circle. In addition, a circle can be formed by the intersection of a cone and a plane that is perpendicular to the axis of the cone:

In a rectangular coordinate plane, where the center of a circle with radius r is (h,k), we have

Calculate the distance between (h,k) and (x,y) using the distance formula,

(x−h)2+(y−k)2=r

Squaring both sides leads us to the equation of a circle in standard formThe equation of a circle written in the form (x−h)2+(y−k)2=r2 where (h,k) is the center and r is the radius.,

(x−h)2+(y−k)2=r2

In this form, the center and radius are apparent. For example, given the equation (x−2)2+ (y + 5)2=16 we have,

(x−h)2+ (x−k)2=r2↓↓↓(x−2)2+[y−(−5)]2=42

In this case, the center is (2,−5) and r=4. More examples follow:

Equation	Center	Radius
(x−3)2+(y−4)2=25	(3,4)	r=5
(x−1)2+(y+2)2=7	(1,−2)	r=7
(x+4)2+(y−3)2=1	(−4,3)	r=1
x2+(y+6)2=8	(0,−6)	r=22

The graph of a circle is completely determined by its center and radius.

Example 1

Graph: (x−2)2+(y+5)2=16.

Solution:

Written in this form we can see that the center is (2,−5) and that the radius r=4 units. From the center mark points 4 units up and down as well as 4 units left and right.

Then draw in the circle through these four points.

Answer:

As with any graph, we are interested in finding the x- and y-intercepts.

Example 2

Find the intercepts: (x−2)2+(y+5)2=16.

Solution:

To find the y-intercepts set x=0:

(x−2)2+(y+5)2=16(0−2)2+(y+5)2=164+(y+5)2=16

For this equation, we can solve by extracting square roots.

(y+5)2=12y+5=±12y+5=±23y=−5±23

Therefore, the y-intercepts are (0,−5−23) and (0,−5+23). To find the x-intercepts set y=0:

(x−2)2+(y+5)2=16(x−2)2+(0+5)2=16(x−2)2+25=16(x−2)2=−9x−2=±−9x=2±3i

And because the solutions are complex we conclude that there are no real x-intercepts. Note that this does make sense given the graph.

Answer: x-intercepts: none; y-intercepts: (0,−5−23) and (0,−5+23)

Given the center and radius of a circle, we can find its equation.

Example 3

Graph the circle with radius r=3 units centered at (−1,0). Give its equation in standard form and determine the intercepts.

Solution:

Given that the center is (−1,0) and the radius is r=3 we sketch the graph as follows:

Substitute h, k, and r to find the equation in standard form. Since (h,k)=(−1,0) and r=3 we have,

(x−h)2+(y−k)2=r2[x−(−1)]2+(y−0)2=32(x+1)2+y2=9

The equation of the circle is (x+1)2+y2=9, use this to determine the y-intercepts.

(x+1)2+y2=9  Set x=0 to and solve for y.(0+1)2+y2=91+y2=9y2=8y=±8y=±22

Therefore, the y-intercepts are (0,−22) and (0,22). To find the x-intercepts algebraically, set y=0 and solve for x; this is left for the reader as an exercise.

Answer: Equation: (x+1)2+y2=9; y-intercepts: (0,−22) and (0,22); x-intercepts: (−4,0) and (2,0)

Of particular importance is the unit circleThe circle centered at the origin with radius 1; its equation is x2+y2=1.,

x2+y2=1

Or,

(x−0)2+(y−0)2=12

In this form, it should be clear that the center is (0,0) and that the radius is 1 unit. Furthermore, if we solve for y we obtain two functions:

x2+y2=1y2=1−x2y=±1−x2

The function defined by y=1−x2 is the top half of the circle and the function defined by y=−1−x2 is the bottom half of the unit circle:

Try this! Graph and label the intercepts: x2+(y+2)2=25.

Answer:

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