Example #5: - Mrs. Yu's Website · Web viewGeometry Unit 2 Lesson 8 and 9 Lesson 8 Equation of A circle Remember that a circle is the set of points (x, y) in a plane that are equidistant

Geometry Unit 2 Lesson 8 and 9

Lesson 8 Equation of A circle

Remember that a circle is the set of points (x, y) in a plane that are equidistant

from a given point (h, k), which is called the center. If we know the coordinates of the center (h, k) and the length of the radius r, we

can use the distance formula, because the radius is the distance from the center to any point (x, y) on the circumference of the circle.

Let (h, k) be the coordinates of the center and the radius equal to r where (x, y) are the coordinates of any point on the circle.

This might help!When deriving the equation of a circle from the distance formula, let (x1, y1) be any point (x, y) on the circumference of the circle and (x_, y2) be the center point (h, k) of the circle.We want to find the distance from the center of the circle (h, k) to any point on the circle (x, y). That distance is the radius, r. The distance formula can now be written:

If we square both sides of the equation, we obtain the following:

We call this version of the distance formula the equation of a circle written in standard form.

THEOREM 9-2 The formula for a circle is (x - h) 2 + (y - k) 2 = r 2 , where (h, k) are the

  coordinates of the center and r is the length of the radius.

Corollary: The formula for a circle whose center is the origin is x 2 + y 2 = r 2.

Example #4:Make note!When writing the equation of a circle, make sure to pay attention to negative values for h or k!

Write the equation of the circle with center at (4, -5) and radius of 3. Determine algebraically whether or not the following points are on the circumference of the circle: (4, -2) and (5, 1).

Solution:Write the equation of the circle:

In example #4, we used the equation of a circle centered at (h, k) with a radius of length  r to prove algebraically whether or not a point exists on the circumference of a circle. The equation of a circle, along with the center point (h, k) and the length of the radius r can also be used to determine if a point is on the interior or exterior of the circle.

If a point is in the interior of the circle, it will satisfy the inequality(x - h) 2 + (y - k) 2 < r 2.

If a point is in the exterior of the circle, it will satisfy the inequality(x - h) 2 + (y - k) 2 > r 2.

Example #5:Given the circle with the equation (x - 2) 2 + (y + 3) 2 = 25, determine each of the point’s location with respect to the circle: A(2, 1), B(-2, 5), and C(2, 2).

Solution:A (2, 1) B (-2, 5) C(2, 2)

Let x=2, y=1: Let x= -2, y=5:  Let x=2, y=2: 

Point A is located on the interior of the circle.

Point B is located on the exterior of the circle.

Point C is located on the circumference of the circle.

Using Pythagorean Theorem with Circles

Remember that the Pythagorean Theorem states that for any right triangle, a2 + b2 = c2, where a and b are the lengths of the sides of the triangles and c is the length of the hypotenuse. Using this formula, you can find a missing side if you know the lengths of the other two sides.

Start by drawing a right triangle within a circle using the radius as the hypotenuse.

The center of the circle is at (h, k). The right triangle’s side lengths are x - h and y - k, and the hypotenuse is r.

Using these values in the Pythagorean Theorem gives (x - h)2 + (y - k)2 = r2. Since this equation holds for any point P on the circle, it is the equation of the circle.

Example #6:

Write the equation of a circle with a center at (2, 3) and the point (5, 7) on the circumference of the circle.

Solution:First, use the given points to find the length of the radius of the circle.

Let (2, 3) = (h, k) and (5, 7) = (x, y). Substitute the values of h, k, x, and y into the equation of a circle and solve for r.

(5 - 2)2 + (7 - 3)2 = r2

(3)2 + (4)2  (4)2 = r2

25 = r2

r = 5Write the equation of the circle with the given center at (2, 3) and  r = 5.

Standard and General Form of a Circle

Complete the Square

Example #8:Rewrite x2 + y2 + 8x - 14y + 40 = 0 in standard form. Find the center and the radius of the circle.

Solution:When completing the square on the general form of a circle, you will complete the square on both of the squared variables, x2 and y2.

x2 + y2 + 8x - 14y + 40 = 0

Step 1: use inverse operations to move any constant term to the opposite side of the equation

x2 + y2 + 8x - 14y= -40

Step 2: rearrange the left side of the equation so that like terms are placed in descending order

x2 + 8x + y2  - 14y= -40

Step 3: separate the polynomial into two sets of binomials with like terms

(x2 + 8x) + (y2  - 14y)= -40

Step 4: prepare the equation for completing the square and writing the equation of a circle in standard form

Step 5: complete the square on both variables x2 and y2

Step 6: combine like terms on the right side of the equation

(x = 4)2 + (y - 7)2 = 25

Step 7: identify the center and the radius of the circle

center (-4, 7)radius = 5

Lesson 9 Midpoint Formula

MIDPOINT FORMULA: [ (x2 + x1),  (y2 + y1)] OR 

Here’s why it works:

Horizontal:

First, find the difference of the two x-coordinates:

P1P2 = x2 - x1

Next, take half of that distance:

P1M =  (x2 - x1)

Now, to find the midpoint add ½ the distance to the first x-coordinate:

x3 = x1 +   (x2 - x1)

Now, use the Distributive Property:

x3= x1 -  x1 +  x2

Combine like terms to obtain:

x3 =  x1 +  x2

Now, factor the result:

x3 =  (x1 + x2)                                                             

The x-coordinate of the midpoint is   the sum of the x-coordinates of the endpoints. The y-coordinate of all three points is the same.

Vertical:

The process to find the midpoint of a vertical segment is similar, except we use the y-coordinates.

P1P2 = y2 - y1

P1M =  (y2 - y1)

y3 = y1 +  (y2 - y1)

= y1 +  y2 -  y1

=  y2 +  y1

y3 =  (y2 + y1)

The y-coordinate of the midpoint is   the sum of the y-coordinates of the endpoints. The x-coordinates of all three points are the same.

Finding the midpoint of a slanting line:

The method for finding the midpoint of a slanting line may not be as apparent as the methods for horizontal or vertical lines, but it is actually just as easy. The x-coordinate of R, the midpoint of P1P3, is   (x2 + x1). The y-coordinate of S, the midpoint of P2P3, is   (y2 + y1). The x-coordinate of M, the midpoint of P1P2, is the same as the x-coordinate of R,  (x2 + x1 ). The y-coordinate of M is the same as the y-coordinate of S,  (y2 + y1). Therefore, the coordinates of the midpoint are given by Theorem 9-3.

THEOREM 9-3The coordinates of the midpoint of the segment connecting points (xl, y1) and (x2, y2) are [ (x2 + x1),    (y2 + y1)].

To find the x-coordinate of the midpoint of a slanted segment, you simply add the x-coordinates together and then divide by two. To find the y-coordinate of the midpoint, add the two y-coordinates and divide by 2.

Now that you have found and used the midpoint in a variety of different ways, you can prove it. When writing a proof, you must be very specific and detailed. Proofs require proof. In other words, you need evidence from theorems, properties, and definitions to support your claims, or statements.

Example #2:The following is a step-by-step justification proving the value of x.

Solution:

Given: M is the midpoint of AB;

AM = 7x + 1; MB = 10x – 8

Prove: x = 3

Statements Reasons

1. M is the midpoint of AB;AM = 7x + 1; MB = 10x – 8

Given

2. AM = MB 2. Definition of midpoint

3. 7x + 1 = 10x – 8 3. Substitution

4. -3x + 1 = - 8 4. Subtraction Property of Equality

5. -3x = - 9 5. Subtraction Property of Equality

6. x = 3 6. Division Property of Equality

Notice that each statement has a reason that is a definition or property to support it. When writing proofs, this is required. You will have to recall information from previous

lessons and units to help fill in your proofs. If you can’t support it with evidence, you can’t use it. Make sure every reason justifies what you are stating. Remember, if it helps to draw an illustration, feel free to do so.