Notes 10.2: Circles

The parabola is one of a family of curves called conic sections. Conic sections are formed by the intersection of a double right cone and a plane. There are four types of conic sections: circles, ellipses, hyperbolas, and parabolas.

Although the parabolas you studied so far are functions, most conic sections are not. This means that you often must use two functions to graph a conic section on a calculator. We will review how to graph conics on a calculator at the end of the lesson.

I. Parabolas

A circle is the set of points in a plane that are a fixed distance, called the radius, from a fixed point, called the center. Because all of the points on a circle are the same distance from the center of the circle, you can use the Distance Formula to find the equation of a circle.

II. CirclesA. Definition of Circle

Ex 1: Write the equation of a circle with center

(–3, 4) and radius r = 6.

Standard Form of the Equation of a Circle

Use the Distance Formula with (x2, y2) = (x, y), (x1, y1) = (–3, 4), and distance equal to the radius, 6.

Use the Distance Formula.

Substitute.

Square both sides.

Notice that r2 and the center are visible in the equation of a circle. This leads to a general formula for a circle with center (h, k) and radius r.

If the center of the circle is at the origin, the equation simplifies to x2 + y2 = r2.

Helpful Hint

STANDARD FORM OF THE EQUATION OF A CIRCLE

Ex 2a: Write the equation of the circle, in standard form, with center (0, 6) and radius r = 1

(x – 0)2 + (y – 6)2 = 12

x2 + (y – 6)2 = 1

(x – h)2 + (y – k)2 = r2 Equation of a circle

Substitute.

Use the Distance Formula to

find the radius.

Substitute the values into the

equation of a circle.

(x + 4)2

+ (y – 11)2

= 225

(x + 4)2

+ (y – 11)2

= 152

Ex 2b: Write the equation of the circle, in standard form, with center (–4, 11) and containing the point (5, –1).

GENERAL FORM: x2 + y2 + Dx + Ey + F= 0

Each form has its advantages. For example standard form is great for determining the center and radius with only a glance at the equation. General form is better for substituting and testing ordered pairs. Due to the fact that we use both of these forms, we must be able to interchange from one form to another. We will first transform from standard form to general form.

III. General FormA. Form and Uses

Ex 3: Transform (x – 3)2 + (y + 5)2 = 64 to general form.

(x – 3)2 + (y + 5)2 = 64

B. Standard to General

Multiply the binomials and rearrange until you get to General form.

(x – 3) (x – 3) + (y + 5) (y + 5) = 64

(x2 – 3x – 3x + 9) + (y2 + 5y + 5y + 25) = 64

x2 – 6x + 9 + y2 + 10y + 25 = 64

x2 + y2 – 6x + 10y + 9 + 25 – 64 = 0

x2 + y2 – 6x + 10y – 30 = 0

x2 + y2 – 6x + 10y – 30 = 0 … General Form

C. General to StandardTo convert from general form to standard, we must complete the square, making sure to keep the equation balanced. Remember that completing the square means creating Perfect Square Trinomials (PSTs).

Some examples of perfect square trinomials are:

Notice that the sign of the middle term can be positive or negative. There is a relationship between the coefficient of the middle term and the last term:

x2 + 2x + 1

x2 - 4x + 4

x2 + 6x + 9

x2 + 8x + 16

x2 - 10x + 25

x2 + 12x + 36

x2 + 14x + 49

x2 - 16x + 64

x2 + 18x + 81

x2 - 20x + 100x2 - 20x + 100

-4 4

14 49

-20 100

42

42

492

142

1002

202

x2 + 2x + 1

x2 - 4x + 4

x2 + 14x + 49

𝑏

2

2

x2 + 6x + 9 = (x + 3)(x + 3) = (x + 3)2

x2 + 8x + 16 = (x + 4)(x + 4) = (x + 4)2

x2 + 2x + 1 = (x + 1)(x + 1) = (x + 1)2

x2 - 4x + 4= (x - 2)(x - 2) = (x - 2)2

x2 - 20x + 100 = (x - 10)(x - 10) = (x - 10)2

x2 + 32x + 256 = (x + 16)(x + 16) = (x + 16)2

When we factor PSTs, we get two identical binomial factors.

x2 + 7x + = (x + )(x + ) = (x + )24

492

7

2

7

2

7

Ex 4a: Find the center and radius of the circle whose equation is: x2 + y2 – 8x + 4y + 11 = 0.

To determine the center and radius, we must transform general form of the equation to standard form.

x2 + y2 – 8x + 4y + 11 = 0

x2 – 8x + y2 + 4y + 11 = 0

(x2 – 8x + ) + (y2 + 4y + ) = -11

We must complete the square for each trinomial by adding the appropriate values. If we add those values to the left side of the equation, we must add the same values to the right side in order to keep both sides of the equation equal.

(x2 – 8x + 16) + (y2 + 4y + 4) = -11 + 16 + 4

(x – 4)2 + (y + 2)2 = 9

Centrer(4,-2) Radius = 3

Put the x’s and y’s together

Create a space for the PST

Factor

Ex 5: Graph each equation on a graphing calculator. Identify each conic section. Then describe the center and intercepts.

IV. Graphing Circles on a Graphing Calculator

(x – 1)2

+ (y – 1)2

= 1

Step 1 Solve for y so that the expression can be used in a graphing calculator.

Subtract (x – 1)2

from both sides.(y – 1)2 = 1 – (x – 1)

2

Take square root of both sides.

Then add 1 to both sides.

Step 2 Use two equations to see the complete graph.

Use a square window on your

graphing calculator for an accurate

graph. The graphs meet and form a

complete circle, even though it might

not appear that way on the calculator.

Check Use a table to confirm the intercepts.

The graph is a circle with center (1, 1)

and intercepts (1,0) and (0, 1).

A tangent is a line in the same plane as the circle that intersects the circle at exactly one point. Recall from geometry that a tangent to a circle is perpendicular to the radius at the point of tangency.

IV. Tangents

Ex 6: Write the equation of the line tangent to the circle x2 + y2 = 29 at the point (2, 5).

Step 1 Identify the center and radius of the circle.

From the equation x2

+ y2

= 29, the circle has

center of (0, 0) and radius r = .

Example 6 Continued

Step 2 Find the slope of the radius at the point of tangency and the slope of the tangent.

Substitute (2, 5) for (x2 , y2 )

and (0, 0) for (x1 , y1 ).

Use the slope formula.

The slope of the radius is .5

2

Because the slopes of perpendicular lines are negative reciprocals, the slope of the tangent is .2

5–

Example 6 Continued

Use the point-slope formula.

Rewrite in slope-intercept form.

Substitute (2, 5) (x1 , y1 ) and – for m. 2

5

Step 3 Find the slope-intercept equation of the tangent by using the point (2, 5) and the slope m = . 2

5–

Example 6 Continued

The equation of the line that is tangent to

x2

+ y2

= 29 at (2, 5) is .

Check Graph the

circle and the line.