General Equation of a Circle

Which of the following equations of circle in general form given the center at (3, 4) and radius of 5 units?

• (x – 3)² - (y – 4)² = 25

• (x – 3)² + (y – 4)² = 25

• x² + y² + 6x - 8y - 25 = 0

• x² + y² - 6x - 8y = 0

Which of the following equations of circle in general form given the center at (5, -8) and radius of 11 units?

• x² + y² -10x + 16y + 32 = 0

• x² + y² -10x + 16y - 32 = 0

• (x - 5)² + (y + 8)² = 11

• (x - 5)² + (y + 8)² = 121

How can we find the general form of a circle? 

• divide out the center-radius form
• multiply out the center-radius form
• factor out the center-radius form
• add up the center-radius form

In order to convert from general form to center-radius form, you must ___________. 

• Multiply out
• expand
• factor
• complete the square

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

Determine the equation of the circle in standard form: x² + y² + 6x - 12y + 20 = 0

• (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 equation of a circle with center (5,-3) and radius of 9

• x²+y²-10x+6y-47=4
• x²+y²-10x-6y-47=0
• 2x²+y²+10x+6y-47=0
• x²+y²-10x+6y-47=0

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 equation of a circle with center (5,-3) and radius of 9

• x²+y²-10x+6y-47=4
• x²+y²-10x-6y-47=0
• 2x²+y²+10x+6y-47=0
• x²+y²-10x+6y-47=0

Tags

CCSS.HSG.GPE.A.1