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________________________________, Note: Please answer my question correctly, nonsense answer will be reported. ________________________________ , Lesson: Illustrating the Center-Radius of the Equation of a Circle ________________________________

Written by Winona · 49 sec read >


________________________________


Note: Please answer my question correctly, nonsense answer will be reported. ________________________________

Lesson: Illustrating the Center-Radius of the Equation of a Circle ________________________________​

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✒️CIRCLE EQUATIONS

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 \largenderline{\mathbb{ANSWER}:}

Part A:

 \qquad \large \rm 1) \; (x - 2)^2 + (y + 4)^2 = 25

 \qquad \large \rm 2) \; (x - 3)^2 + y^2 = 1/9

 \qquad \large \rm 3) \; x^2 + y^2 = 1/16

 \qquad \large \rm 4) \; (x + 7)^2 + (y - 1)^2 = 5

 \qquad \large \rm 5) \; (x + 6)^2 + (y + 3)^2 = 48

 \qquad \large \rm 6) \; x^2 + (y - 6)^2 = 81

Part B:

 \qquad \large \rm 7) \; (x - 3)^2 + (y + 4)^2 = 45

 \qquad \large \rm 8) \; (x - 5)^2 + (y -  1)^2 = 18

 \qquad \large \rm 9) \; x^2 + y^2 = 25

 \qquad \large \rm 10) \; x^2 + y^2 = 16

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 \largenderline{\mathbb{SOLUTION}:}

For Part A and B: The equation of the circle in standard form is written as:

  •  (x - h)^2 + (y - k)^2 = r^2

Where (h,k) is the center and r is the radius. Substitute each given to get its equation.

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“Part A”

Number 1:

  •  (x - 2)^2 + \bigy - (\text-4)\big^2 = 5^2
  •  (x-2)^2 + (y + 4)^2 = 25

Number 2:

  •  (x - 3)^2 + (y - 0)^2 = (1/3)^2
  •  (x - 3)^2 + y^2 = 1/9

Number 3:

  •  (x - 0)^2 + (y - 0)^2 = (1/4)^2
  •  x^2 + y^2 = 1/16

Number 4:

  •  \bigx - (\text-7)\big^2 + (y - 1)^2 = (\sqrt5\,)^2
  •  (x + 7)^2 + (y - 1)^2 = 5

Number 5:

  •  \bigx - (\text-6)\big^2 + \bigy - (\text-3)\big^2 = (4\sqrt3\,)^2
  •  (x + 6)^2 + (y + 3)^2 = 16(3)
  •  (x + 6)^2 + (y + 3)^2 = 48

Number 6:

  •  (x - 0)^2 + (y -6)^2 = 9^2
  •  x^2 + (y - 6)^2 = 81

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“Part B”

Number 7:

Substitute the given center in the standard form of the equation.

  •  (x - 3)^2 + \bigy - (\text-4)\big^2 = r^2
  •  (x - 3)^2 + (y + 4)^2 = r^2

Find the square of the radius if it passes through (6,2)

  •  (6 - 3)^2 + (2 + 4)^2 = r^2
  •  (3)^2 + (6)^2 = r^2
  •  9 + 36 = r^2
  •  45 = r^2

Substitute the square of the radius to the equation.

  •  (x - 3)^2 + (y + 4)^2 = 45

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Number 8:

Substitute the given center in the standard form of the equation.

  •  (x - 5)^2 + (y - 1)^2 = r^2

Find the square of the radius if it passes through (8,-2)

  •  (8 - 5)^2 + (\text-2 - 1)^2 = r^2
  •  (3)^2 + (\text-3)^2 = r^2
  •  9 + 9 = r^2
  •  18 = r^2

Substitute the square of the radius to the equation.

  •  (x - 5)^2 + (y - 1)^2 = 18

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Number 9:

Substitute (0,0) as the given center in the standard form of the equation since it is at the origin.

  •  (x - 0)^2 + (y - 0)^2 = r^2
  •  x^2 + y^2 = r^2

Find the square of the radius if it passes through (4,3)

  •  4^2 + 3^2 = r^2
  •  16 + 9 = r^2
  •  25 = r^2

Substitute the square of the radius to the equation.

  •  x^2 + y^2 = 25

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Number 10:

Find the midpoint between the endpoints because that that would be the center of the circle.

 \begin{align} & \bold{Formula:} \\ & \quad \boxed{\rm Midpoint = \bigg(\frac{x_2+x_1}2,\,\frac{y_2+y_1}2\bigg)} \end{align}

  •  \rm Center = \bigg(\frac{\text-4 + 4}2,\, \frac{0+0}2 \bigg) \\
  •  \rm Center = \bigg(\frac{\,0\,}2,\, \frac{\,0\,}2 \bigg) \\
  •  \rm Center = (0,0)

Thus, the center is at the origin. Substitute it in the standard form of the equation.

  •  (x - 0)^2 + (y - 0)^2 = r^2
  •  x^2 + y^2 = r^2

Find the square of the radius if it passes through one of the given endpoints of the diameter: (4,0)

  •  4^2 + 0^2 = r^2
  •  16 + 0 = r^2
  •  16 = r^2

Substitute the square of the radius to the equation.

  •  x^2 + y^2 = 16

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