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Equations of Circles

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Equations of Circles

What type of graph is produced by the equation x^2-2x+y^2-4y+1=0 ? How can you turn this equation into graphing form in order to graph it?

Watch This

http://www.youtube.com/watch?v=g1xa7PvYV3I Conic Sections: The Circle

Guidance

A circle is one example of a conic section. A circle is the set of all points that are equidistant from a center point. The general equation of a circle is  (x-h)^2+(y-k)^2=r^2 where  (h, k) is the center of the circle and  r is the radius of the circle.

You can derive the equation of a circle with the help of the Pythagorean Theorem.

Example A

Label the three sides of the right triangle below in terms of x, y , and r . Then, use the Pythagorean Theorem to write an equation that shows the relationship between the three sides.

Solution: The three sides of the triangle are x , y , and r , where  r is the hypotenuse.

By the Pythagorean Theorem, x^2+y^2=r^2 .

Example B

Do all points on the circle from Example A satisfy the equation x^2+y^2=r^2 ?

Solution: Yes, anywhere you move the point (x, y) , the lengths of the sides of the triangle will still be  xy , and r , so  x^2+y^2=r^2 will still be true.

In fact, the set of points that make up the circle is exactly the set of points that satisfy the equation x^2+y^2=r^2 . This is why the general equation of a circle centered at the origin is x^2+y^2=r^2 , where  r is the radius of the circle.

Example C

Why does a circle with center  (h, k) and radius  r have the general equation (x-h)^2+(y-k)^2=r^2 ?

Solution: Pick a point on the circle and draw a right triangle connecting that point with the center of the circle.

The three sides of the triangle have lengths x-hy-k and r .

The relationship between the sides of the triangle is shown with the Pythagorean Theorem:

(x-h)^2+(y-k)^2=r^2

Points on the other side of the circle produce triangles with slightly different side lengths:

However, the relationship between the three sides of this triangle is the same, as shown below:

(h-x)^2+(k-y)^2 &= r^2 \\ ((-1)(x-h) )^2+((-1)(y-k) )^2 &= r^2 \\(-1)^2 (x-h)^2+(-1)^2 (y-k)^2 &= r^2 \\(x-h)^2+(y-k)^2 &= r^2

The set of points that satisfy the equation  (x-h)^2+(y-k)^2=r^2 are the set of points that make up the circle.

Concept Problem Revisited

The equation  x^2-2x+y^2-4y+1=0 is the equation for a circle. You can tell it is a circle because there is both an  x^2 term and a  y^2 term, and the coefficients of each are the same. To turn this equation into the form (x-h)^2+(y-k)^2=r^2 , you must complete the square twice. Remember that when completing the square, you are trying to figure out what number to add to make a perfect square trinomial. In order to maintain the equality of the equation, you must add the same numbers to both sides of the equation.

& x^2-2x+ \underline{\; \; \; \; \; \; \; \;}+y^2-4y+\underline{\; \; \; \; \; \; \; \;}=-1 \\& x^2-2x+ \underline{1} +y^2-4y+ \underline{4} =-1 \underline{+1+4} \\& x^2-2x+1+y^2-4y+4=4 \\& (x-1)^2+(y-2)^2=4

Vocabulary

A circle is the set of all points equidistant from a given point. The general equation of a circle is (x-h)^2+(y-k)^2=r^2 .

Guided Practice

1. What is the equation of a circle centered at  (-4, 5) with a radius of \sqrt{2} ?

2. What are the center and radius of the circle described by the equation:

x^2-4x+y^2+10y+13=0?

3. Graph the circle from #2.

Answers:

1. h=-4 , k=5 , r=\sqrt{2} . The equation of the circle is (x-(-4) )^2+(y-5)^2= \left ( \sqrt{2} \right )^2 . This can be simplified to (x+4)^2+(y-5)^2=2 . Notice that the signs of the 4 and the 5 in the equation are opposite from the signs of the 4 and the 5 in the center point.

2. Complete the square to rewrite the equation in graphing form:

x^2-4x+\underline{4}+y^2+10y+\underline{25} &=-13 \underline{+4+25} \\(x-2)^2+(y+5)^2 &=16

The center of the circle is  (2, -5) and the radius is 4.

3.

 

Practice

Graph the following circles:

1. (x+2)^2+(y-5)^2=1

2. (x-1)^2+(y+1)^2=3

3. (y+4)^2+(x-3)^2=\frac{1}{4}

4. x^2+(y-6)^2=8

5. (x-2)^2+y^2=25

Find the center and radius of the circle described by each equation.

6.  x^2+4x+y^2+4y+7=0

7. x^2-2x+y^2+12y+32=0

8. x^2-2x+y^2-2y=0

9. x^2+8x+y^2+9=0

10. x^2+2x+y^2-6y+1=0

11. Explain why a circle with radius  r that is centered at the origin will have the equation x^2+y^2=r^2 .

12. Use the Pythagorean Theorem to help explain why a circle with a center in the second quadrant with radius  r and center  (h, k) will have the equation (x-h)^2+(y-k)^2=r^2 . Note that in the second quadrant the value for   h will be negative and the value for   k will be positive.

13. Use the Pythagorean Theorem to help explain why a circle with a center in the third quadrant with radius  r and center  (h, k) will have the equation (x-h)^2+(y-k)^2=r^2 . Note that in the third quadrant the value for  h will be negative and the value for   k will be negative.

14. Write the equation of the following circle:

15. Write the equation of the following circle:

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