Lesson Check Two: Equations of a Circle (Harder Problems)

In the equation (x-4)²+(x-3)²=25, the center of the circle is at...

• (-3, -4)
• (4, 3)
• (-4, -3)
• (3, 4)

Tags

CCSS.HSG.GPE.A.1

The previous problem was in the form: (x-h)^2 + (y-k)^2 = r^2.  When an equation is in this form it is easy to find the center and the radius.  For example, what is the radius of  $\left(x-9\right)^2+\left(y+2\right)^{2\ }=\ \ 81$  

• 81

• 9

• 3

• -9

If given an equation that looks like this: x² + y² + 14x - 2y = -1, you cannot easily see the center and the radius. Because of that you must manipulate the equation INTO the form you want. What would be the first step?

• Get the x's together and get the y's together.

• Factor

• Add (b/2)^2 to both sides.

• Add 1 to the right hand side.

Determine the center and radius: x² + y² + 14x - 2y = -1

• center= (-7,2) radius= 3
• center= (-7,1) radius= 7
• center= (-4,1) radius= 5

Tags

CCSS.HSG.GPE.A.1

Determine the center and radius:
x² + y² - 10x + 8y -8 = 0

• C = (5,-4) r = 7
• C = (-5,4) r = 7
• C = (-4,5) r = 49
• C = (5,-4) r = 49

Tags

CCSS.HSG.GPE.A.1

What is the center for the circle with this equation x²+y²+4x+6y-36=0?

• (-2,5)

• (2,7)

• (-2,-3)

Tags

CCSS.HSG.GPE.A.1

What is the radius of the given circle:

x² + y² - 10x + 8y - 23 = 0?

• 4

• 2

• 8

What is the center of the circle with the equation x² + y² + 4x + 6y - 36 = 0 ?

• (-2,5)

• (2,7)

• (-2,-3)

Determine the center of the circle.

x² + y² + 10x - 8y - 8 = 0

• C = (5,-4)

• C = (-5,4)

• C = (-4,5)

• C = (5,-4)

We saw that the first step was to put the x's together and put the y's together like so: x² + 14x + y² - 2y = -1. After that we should ___________ ____ ___________ for the x's and for the y's.

• Complete the Square

• Multiply by the Reciprocal

• Multiply it Out

• Factor the GCF