SECTION 10.8Equations of Circles

Example 1: a) Write the equation of the circle with a center at (3, –3) and a radius of 6.    

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

(x – 3)2 + (y – (–3))2 = 62 Substitution

(x – 3)2 + (y + 3)2 = 36 Simplify.

Answer: (x – 3)2 + (y + 3)2 = 36

Example 1:

b) Write the equation of the circle graphed below.

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

(x – 1)2 + (y – 3)2 = 22 Substitution

(x – 1)2 + (y – 3)2 = 4 Simplify.

Answer: (x – 1)2 + (y – 3)2 = 4

Example 1:

c) Write the equation of the circle graphed below.

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

(x – 0)2 + (y – 3)2 = 32 Substitution

x2 + (y – 3)2 = 9 Simplify.

Answer: x2 + (y – 3)2 = 9

Example 2: a) Write the equation of the circle that has its center at (–3, –2) and passes through (1, –2). 

2 2

2 1 2 1

2 2

1 1

2 2

Distance Formula

1 3 2 2 , 3, 2 and

, 1, 2

16 Simplify

4

r x x y y

x y

x y

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

(x – (–3))2 + (y – (–2))2 = 42 Substitution

(x + 3)2 + (y + 2)2 = 16 Simplify.

Answer: (x + 3)2 + (y + 2)2 = 16

Example 2: b) Write the equation of the circle that has its center at (–1, 0) and passes through (3, 0).

2 2

2 1 2 1

2 2

1 1

2 2

Distance Formula

1 3 0 0 , 1,0 and

, 3,0

16 Simplify

4

r x x y y

x y

x y

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

(x – (–1))2 + (y – 0)2 = 42 Substitution

(x + 1)2 + y2 = 16 Simplify.

Answer: (x + 1)2 + y2 = 16

Example 3: a) The equation of a circle is x2 – 4x + y2 + 6y = –9. State the coordinates of the center and the measure of the radius. Then graph the equation.

x2 – 4x + y2 + 6y= –9

Original equation

x2 – 4x + 4 + y2 + 6y + 9 = –9 + 4 + 9 Complete the squares.

(x – 2)2 + (y + 3)2

= 4 Factor and simplify.

(x – 2)2 + [y – (–3)]2

= 22 Write +3 as – (–3) and 4 as 22.

(x – 2)2 + [y – (–3)]2 = 22

(x – h)2 + [y – k]2 = r2

Answer: So, h = 2, y = –3, and r = 2. The center is at (2, –3), and the radius is 2.

a) b)

c) d)

Example 3:

b) Which of the following is the graph of x2 + y2 – 10y = 0?

Example 4: a) Strategically located substations are extremely important in the transmission and distribution of a power company’s electric supply. Suppose three substations are modeled by the points D(3, 6), E(–1, 1), and F(3, –4). Determine the location of a town equidistant from all three substations, and write an equation for the circle.

SKIP!!!!!!!

 

Example 4:

b) The designer of an amusement park wants to place a food court equidistant from the roller coaster located at (4, 1), the Ferris wheel located at (0, 1), and the boat ride located at (4, –3). Determine the location for the food court. 

SKIP!!!!!!!

Example 5: a) Find the point(s) of intersection between x2 + y2 = 32 and y = x + 8.

Graph these equations on the same coordinate plane.

There appears to be only one point of intersection. You can estimate this point on the graph to be at about (–4, 4). Use substitution to find the coordinates of this point algebraically.

x2 + y2 = 32 Equation of circle.

x2 + (x + 8)2 = 32Substitute x + 8 for y.

x2 + x2 + 16x + 64 = 32Evaluate the square.

2x2 + 16x + 32 = 0 Simplify.

x2 + 8x + 16 = 0 Divide each side by 2.

(x + 4)2= 0 Factor.

x = –4 Take the square root of each side.

Use y = x + 8 to find the corresponding y-value.

(–4) + 8 = 4

The point of intersection is (–4, 4).

Answer: (–4, 4)

Example 5: b) Find the points of intersection between x2 + y2 = 16 and y = –x. x2 + y2 = 16 Equation of

circle.

x2 + (–x)2 = 16 Substitute –x for y.

x2 + x2 = 16 Evaluate the square.

2x2 = 16 Simplify.

x2 = 8 Divide each side by 2.

x = Take the square root of each side.

2 2

2 2, 2 2 and 2 2,2 2

Use y = –x to find the corresponding y-values.