Deriving and Using the 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)

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

• 4
• 2
• 3
• 16

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)

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

See Picture

• A
• B
• C
• D

The diameter of a circle has length 12. The center is at (-5, 2). Give the equation of the circle.

• (x - 2)² + (y + 5)² = 36
• (x - 5)² + (y + 2)² = 6
• (x + 5)² + (y - 2)² = 36
• (x + 2)² + (y - 5)² = 6

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.

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

The diameter of a circle has length 12. The center is at (-5, 2). Give the equation of the circle.

• (x - 2)² + (y + 5)² = 36
• (x - 5)² + (y + 2)² = 6
• (x + 5)² + (y - 2)² = 36
• (x + 2)² + (y - 5)² = 6

What is the equation of a circle with radius 2 and center (3, 4)?

• (x+3)² + (y+4)² = 2
• (x-3)² + (y-4)² = 2
• (x+3)² + (y+4)² = 4
• (x-3)² + (y-4)² = 4