Equation of Circle Standard Form and Complete the Square

Change the following equation to standard form for the equation of a circe: x²+y²-4x+12y-8=0

• (x+2)²+(y-6)²=48

• (x-2)²+(y+6)²=48

• (x-6)²+(y-2)²=48

• (x+2)²+(y-6)²=48

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

Change the following equation to standard form for the equation of a circe: x² + y² + 6x - 12y + = -20.

• (x-3)² + (y+6)² = 25

• (x+3)² + (y-6)² = 25

• (x-3)² + (y+6)² = 45

• (x+3)² + (y-6)² = 5

Tags

CCSS.HSG.GPE.A.1

Determine the equation of the circle in standard form:
x² + y² + 4x + 6y - 36 = 0

• (x+2)² + (y+3)² = 49
• (x+2)² + (y+3)² = 13
• (x-2)² + (y-3)² = 49
• (x-2)² + (y-3)² = 13

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 radius of the given circle:
 x²+y²-10x+8y-23=0?

• 4
• 2
• 8

Tags

CCSS.HSG.GPE.A.1

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

A circle has the equation  x² + y² - 12x + 8y + 3 = 0.  Use completing the square to write the equation in standard form.

• (x - 6)² + (y + 4)² = 49
• (x + 4)² + (y - 6)² = 49
• (x - 6)² + (y + 4)² = 55
• x² + y² = 55

Tags

CCSS.HSG.GPE.A.1