10.7 Circles in the Coordinate Plane

​10.7 Circles in the Coordinate Plane

​" Success is the sum of Small Efforts repeated day in and day out."

R Collier

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Multiple Choice

Write the equation of a circle with center (7, 0) with radius 3. #1

• (x - 7)² + y² = 9

• x² + (y -7)² = 9

• (x - 7)² + y² = 3

• x² + (y -7)² = 3

Multiple Choice

In the equation (x+2)²+(y+3)²=49, the center of the circle is at....#2

• (2, 3)

• (3, 2)

• (-3, -2)

• (-2, -3)

Multiple Choice

#3

• A

• B

• C

• D

Multiple Choice

Which graph corresponds to

(x)²+ (y-6)²=9? #20

• A

• B

• C

• D

Multiple Choice

Which graph corresponds to

(x-4)²+ (y+4)²=25? #21

• A

• B

• C

• D

Multiple Choice

What are the coordinates of the center and the radius of the circle with equation:

(x - 4)² + (y - 3)² = 25 ? #5

• Center (4,3)

  Radius = 5 units

• Center (4,3)

  Radius = 25 units

• Center (-4,-3)

  Radius = 25 units

• Center (-4,-3)

  Radius = 5 units

Multiple Choice

What is the equation of the circle, with center (-2,-7) and radius 3 units? #6

• (x + 2)² + (y + 7)² = 3

• (x - 2)² + (y - 7)² = 9

• (x + 2)² + (y + 7)² = 9

• (x - 2)² + (y - 7)² = 3

Multiple Choice

What is the equation of the circle, with center (-4,5) and radius 6 units? #7

• (x - 4)² + (y + 5)² = 6

• (x - 4)² + (y + 5)² = 36

• (x + 4)² + (y - 5)² = 6

• (x + 4)² + (y - 5)² = 36

Completing the Square

Why do we sometimes need to complete the square?

Multiple Choice

Change the following equation to standard form

x²+y²-4x+12y-8=0. #4

• (x-2)²+(y-6)²=48

• (x-2)²+(y+6)²=48

• (x-6)²+(y-2)²=48

• (x+2)²+(y-6)²=48

Multiple Choice

A circle has the equation x² + y² - 12x + 8y + 3 = 0. Use completing the square to write the equation in standard form. #11

• (x - 6)² + (y + 4)² = 49

• (x + 4)² + (y - 6)² = 49

• (x - 6)² + (y + 4)² = 55

• x² + y² = 55

Multiple Choice

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 and then divide it by 2. The center is simply the midpoint of the given points. $$d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$$   $$Use\ Midpoint\ Formula\ to\ find\ the\ center\ \ \left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$$   #9

• (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

Multiple Choice

The point (3,3) is on a circle whose center is at (-1,3). Write the standard form of the equation of the circle. HINT: USE THE DISTANCE FORMULA TO FIND THE RADIUS $$d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$$   #16

• (x-3)²+(y-3)²=4

• (x-3)²+(y-3)²=16

• (x+1)²+(y-3)²=4

• (x+1)²+(y-3)²=16

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Math Response

What is the equation of the tangent line to the circle at point (3,5)? Write the equation of the line in y-intercept form. #36

Type answer here

Correct Answer: y=-\frac{4}{3}x+9

Deg^°

Rad

Math Response

The line through points B and C is tangent to circle A at point B. What is the equation of the tangent line BC ? #37

Type answer here

Correct Answer: y=-\frac{5}{6}x+\frac{17}{3}

Deg^°

Rad