Distance Formula, Equation of Circle, Midpoint

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

Write the standard equation of the circle with the given center and radius; center (0,0) and radius: 3

• x²+y²=3
• x²+y²=6
• x²+y²=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

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