Distance Formula and Circle Equations

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

The diameter of a circle has endpoints (-8,10) and (-2,4). Find the equation of the circle.

Tags

CCSS.HSG.GPE.A.1

Find the distance between (3, 24) and (7, 56).

• 32.1

• 32.2

• 32.3

• 32.4

Tags

CCSS.HSG.GPE.B.7

Does the point (4,5)  lie on the line 
(x+2)²+(y+2)²=81
Hint substitute x=4 and y=5 and see if the left side is equal the right side.

• yes

• No

Tags

CCSS.HSG.GPE.A.1

Determine whether the  point (5,-3) is on, inside, or outside the circle defined by x²+y²=34

• On

• Inside

• Outside

• Off the graph!

Tags

CCSS.HSG.GPE.A.1

Determine whether the  point (5,-5) is on, inside, or outside the circle defined by x²+y²=34

• On

• Inside

• Outside

• Off the graph!

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

Which graph corresponds to 
(x-8)²+ (y+2)²=

• A

• B

• C

• D

Tags

CCSS.HSG.GPE.A.1

Which graph corresponds to 
(x)²+ (y-6)²=9

• A

• B

• C

• D

Tags

CCSS.HSG.GPE.A.1

Which graph corresponds to 
(x-4)²+ (y+4)²=25

• A

• B

• C

• D

Tags

CCSS.HSG.GPE.A.1