Deriving and using the equation of a circle

The endpoints of a diameter of a circle are (−3,1) and (9,6).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−5)²+(y−7/2)²=169/4
• (x−3)²+(y−3)²=169/4
• (x−3)²+(y−7/2)²=169/4
• (x−7/2)²+(y−7/2)²=169/4

Tags

CCSS.HSG.GPE.A.1

Write the equation of the graphed circle.

• x² + (y + 3)² = 3

• x² + (y – 3)² = 3

• x² + (y + 3)² = 9

• x² + (y – 3)² = 9

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CCSS.HSG.GPE.A.1

Determine the center of the circle given the equation (x+2)^2 + (y-1)^2 = 36

• (3, 4)

• (5, -2)

• (0, 0)

• (-2, 1)

Tags

CCSS.HSG.GPE.A.1

The equation given represents the equation of a circle. What is the radius of the circle?

x² + y² - 4x + 12y +15 = 0

• 6

• 5

• 4

• 7

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CCSS.HSG.GPE.A.1

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CCSS.HSG.GPE.A.1

See Picture

• A

• B

• C

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CCSS.HSG.GPE.A.1

Does the point (-5,6) lie on a circle centered at (-1,3) and has a radius of 5?

• Yes

• No

Tags

CCSS.HSG.GPE.A.1

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CCSS.HSG.GPE.A.1

What is the center and radius of this circle?
(x - 5)² + (y + 2)² = 49

• Center: (5, 2)
   radius = 7
• Center: (5, -2)
   radius = 49
• Center: (5, -2)
   radius = 7
• Center: (-5, -2)
   radius = 7

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