College Algebra Tutorial 29

College Algebra
Answer/Discussion to Practice Problems
Tutorial 29: Circles

WTAMU > Virtual Math Lab > College Algebra > Tutorial 29: Circles

Answer/Discussion to 1a

Center (9, -10) and r = 5

What are the two things we need to write an equation of a circle????
If you said the center and the radius, you are correct.

Looks like we have all the information we need. We are ready to put our equation together.

What value are we going to replace h with?
If you said 9, you are correct!!! h is the first number of the ordered pair of the center of the circle.

What value are we going to replace k with?
If you said -10, you are right on!!! k is the second number of the ordered pair of the center of the circle.

What value are we going to replace r with?
If you said 5, give yourself a pat on the back!!! r is the radius of the circle.

Putting it into standard form we get:

*Plug in 9 for h, -10 for k, and 5 for r
*5 squared is 25
*Standard form of circle with center (9, -10) and radius of 5

Make sure that you are careful when one of your values is negative and you have to subtract it as we did in line 2. y - (-10) is not the same as y - 10.

(return to problem 1a)

Answer/Discussion to 1b

Center (2, 0) and

What are the two things we need to write an equation of a circle????
If you said the center and the radius, you are correct.

Looks like we have all the information we need. We are ready to put our equation together.

What value are we going to replace h with?
If you said 2, you are correct!!! h is the first number of the ordered pair of the center of the circle.

What value are we going to replace k with?
If you said 0, you are right on!!! k is the second number of the ordered pair of the center of the circle.

What value are we going to replace r with?
If you said , give yourself a pat on the back!!! r is the radius of the circle.

Putting it into standard form we get:

*Plug in 2 for h, 0 for k, and sq. root of 2 for r
*sq. root of 2 squared is 2
*Standard form of circle with center (2, 0) and radius sq. root of 2

Generally, when you have an expression where you are subtracting 0, you can simplify it by not writing the - 0. Going from line 2 to 3 above, I simplified the expression y - 0 by writing it in the equivalent form,

(return to problem 1b)

Answer/Discussion to 2a

Step 1: Write the equation of the circle in standard form ().

The equation is already in standard form.

Step 2: Identify the center and radius of circle.

*Standard form of circle

*4 is 2 squared

Make sure that you are careful. The original standard form is , where we have x MINUS h and y MINUS k. So h is the number that we are subtracting from x and k is the number that we are subtracting from y. Also, the right side of the equation is r SQUARED.

So in line 2 above, I rewrote the equation to show what numbers we were actually subtracting on the left side of the equation.

On the right side of the equation I showed what number was being squared. This is to help you see how we get the center (h, k) and the radius (r).

Make sure that you are careful when one of your values is negative and you have to subtract it as we did in line 2. y - (-5) is not the same as y - 5.

What value is h?
If you said 0, you are correct!!! Since there is no value being subtracted from x in the first ( ) that means it is understood that 0 is the number being subtracted, that is the value of h.

What value is k?
If you said -5, you are right on!!! Since -5 is the value being subtracted from y in the second ( ), that is the value of k.

What value is r?
If you said 2, give yourself a pat on the back!!! Since 2 is the number being squared on the right side of the standard form of our equation, that is the value of r.

The center is (0, -5) and the radius is 2.

Step 3: Graph, if needed.

(return to problem 2a)

Answer/Discussion to 2b

Step 1: Write the equation of the circle in standard form ().

Step 1a: Make sure that the coefficients on the and terms is equal to 1.

The coefficients on the

and

terms are both already 1.

Step 1b: Isolate the , x, and y terms.

*Inverse of add. 33 is sub. 33

Step 1c: Complete the square for both x and y.

*Complete the square for both x and y
*Make sure that you add 9 and 49 to BOTH sides

Completing the square for x (as shown above):
b is the coefficient of the -6x term, which in this case is -6. Complete the square by taking 1/2 of b and squaring it. Taking 1/2 of -6 we get -3, and then squaring -3 we get 9.

Completing the square for y (as shown above):
b is the coefficient of the 14y term, which in this case is 14. Complete the square by taking 1/2 of b and squaring it. Taking 1/2 of 14 we get 7, and then squaring 7 we get 49.

Step 1d: Factor both PST's created in step 1c as a binomial squared.

*Factor each PST
*Standard form of circle

Step 2: Identify the center and radius of circle.

*Standard form of circle

*25 is 5 squared

So in line 2 above, I rewrote the equation to show what numbers we were actually subtracting on the left side of the equation.

On the right side of the equation I showed what number was being squared. This is to help you see how we get the center (h, k) and the radius (r).

Make sure that you are careful when one of your values is negative and you have to subtract it as we did in line 2. y - (-7) is not the same as y - 7.

What value is h?
If you said 3, you are correct!!! Since 3 is the value being subtracted from x in the first ( ), that is the value of h.

What value is k?
If you said -7, you are right on!!! Since -7 is the value being subtracted from y in the second ( ), that is the value of k.

What value is r?
If you said 5, give yourself a pat on the back!!! Since 5 is the number being squared on the right side of the standard form of our equation, that is the value of r.

The center is (3, -7) and the radius is 5.

Step 3: Graph, if needed.

(return to problem 2b)

WTAMU > Virtual Math Lab > College Algebra > Tutorial 29: Circles