Equations 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 =6 2 Substitution (x – 3) 2 + (y + 3) 2 = 36Simplify. 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 =2 2 Substitution (x – 1) 2 + (y – 3) 2 = 4Simplify. 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 =3 2 Substitution x 2 + (y – 3) 2 = 9Simplify. Answer: x 2 + (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). (x – h) 2 + (y – k) 2 = r 2 Equation of circle (x – (–3)) 2 + (y – (–2)) 2 =4 2 Substitution (x + 3) 2 + (y + 2) 2 = 16Simplify. 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). (x – h) 2 + (y – k) 2 = r 2 Equation of circle (x – (–1)) 2 + (y – 0) 2 =4 2 Substitution (x + 1) 2 + y 2 = 16Simplify. Answer: (x + 1) 2 + y 2 = 16

Example 3: a) The equation of a circle is x 2 – 4x + y 2 + 6y = –9. State the coordinates of the center and the measure of the radius. Then graph the equation. x 2 – 4x + y 2 + 6y= –9Original equation x 2 – 4x y 2 + 6y + 9 = – Complete the squares. (x – 2) 2 + (y + 3) 2 = 4Factor and simplify. (x – 2) 2 + [y – (–3)] 2 = 2 2 Write +3 as – (–3) and 4 as 2 2.

(x – 2) 2 + [y – (–3)] 2 = 2 2 (x – h) 2 + [y – k] 2 = r 2 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 x 2 + y 2 – 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 x 2 + y 2 = 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. x 2 + y 2 = 32Equation of circle. x 2 + (x + 8) 2 = 32Substitute x + 8 for y. x 2 + x x + 64 = 32Evaluate the square. 2x x + 32= 0 Simplify. x 2 + 8x + 16= 0Divide each side by 2. (x + 4) 2 = 0Factor. x= –4Take 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 x 2 + y 2 = 16 and y = –x. x 2 + y 2 = 16Equation of circle. x 2 + (–x) 2 = 16 Substitute –x for y. x 2 + x 2 = 16 Evaluate the square. 2x 2 = 16 Simplify. x 2 = 8Divide each side by 2. x = Take the square root of each side. Use y = –x to find the corresponding y-values.