Equation of a circle

In the equation (x-3)²+(y-2)²=16, the center of the circle is...

• (3,2)
• (-3, -2)
• (-2, -3)
• (2, 3)

Tags

CCSS.HSG.GPE.A.1

In the equation (x+2)²+(y-3)²=4, the radius of the circle is...

• 4
• 2
• 3
• 16

Tags

CCSS.HSG.GPE.A.1

What is the equation of the circle?

• (x+1)² + (y +1)² = 3
• (x+1)² + (y -1)² = 3
• (x+1)² + (y +1)² = 9
• (x-1)² + (y -1)² = 9

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

What are the coordinates of the center and the radius of the circle with equation:
(x - 4)² + (y - 3)² = 25 ?

• Center (4,3)
  Radius = 5 units
• Center (4,3)
  Radius = 25 units
• Center (-4,-3)
  Radius = 25 units
• Center (-4,-3)
  Radius = 5 units

Tags

CCSS.HSG.GPE.A.1

What is the center of the circle with equation x² + y² = 1?

• (1, 1)
• (0, 0)
• not enough information

Tags

CCSS.HSG.GPE.A.1

What is the equation for this circle?

• (x+1)+(y+1)=9
• x²+y²=9
• x+y=9

Tags

CCSS.HSG.GPE.A.1

Find the equation of the circle with center (−8,3) and (2,−5) is on the circle?
Hint: you need to find the radius by plugging in values for four variables 
x=
y=
h=
k=
in (x-h)²+(y-k)²=r²

• (x+8)²+(y−3)²=164^.5.
• (x+3)²+(y−8)²=164.
• (x)²+(y)²=164.
• (x+8)²+(y−3)²=164.

Tags

CCSS.HSG.GPE.A.1

The endpoints of a diameter of a circle are (3,1) and (9,5).Find the equation of the circle?
Hint: In this problem, we are not given the center or radius however we can find the length of the diameter using the distance formula (Phythagoras) and then divide it by 2. The center is simply the midpoint of the given points.

• (x−6)²+(y−3)²=13
• (x−3)²+(y−6)²=13²
• (x−5)²+(y−9)²=13
• (x−9)²+(y−5)²=169/4

Tags

CCSS.HSG.GPE.A.1

The endpoints of a diameter of a circle are (3,1) and (9,5).Find the equation of the circle?
Hint: In this problem, we are not given the center or radius however we can find the length of the diameter using the distance formula (Phythagoras) and then divide it by 2. The center is simply the midpoint of the given points.

• (x−6)²+(y−3)²=13
• (x−3)²+(y−6)²=13²
• (x−5)²+(y−9)²=13
• (x−9)²+(y−5)²=169/4

Tags

CCSS.HSG.GPE.A.1