Answered

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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 ________________________________​

Note Please Answer My Question Correctly Nonsense Answer Will Be Reported Lesson Illustrating The CenterRadius Of The Equation Of A Circle class=

Sagot :

KAntoineDoixviz

✒️CIRCLE EQUATIONS

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[tex] \large\underline{\mathbb{ANSWER}:} [/tex]

Part A:

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

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

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

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

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

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

Part B:

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

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

[tex] \qquad \large \rm 9) \; x^2 + y^2 = 25 [/tex]

[tex] \qquad \large \rm 10) \; x^2 + y^2 = 16 [/tex]

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[tex] \large\underline{\mathbb{SOLUTION}:} [/tex]

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

  • [tex] (x - h)^2 + (y - k)^2 = r^2 [/tex]

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:

  • [tex] (x - 2)^2 + \big[y - (\text-4)\big]^2 = 5^2 [/tex]

  • [tex] (x-2)^2 + (y + 4)^2 = 25 [/tex]

Number 2:

  • [tex] (x - 3)^2 + (y - 0)^2 = (1/3)^2 [/tex]

  • [tex] (x - 3)^2 + y^2 = 1/9 [/tex]

Number 3:

  • [tex] (x - 0)^2 + (y - 0)^2 = (1/4)^2 [/tex]

  • [tex] x^2 + y^2 = 1/16 [/tex]

Number 4:

  • [tex] \big[x - (\text-7)\big]^2 + (y - 1)^2 = (\sqrt5\,)^2 [/tex]

  • [tex] (x + 7)^2 + (y - 1)^2 = 5 [/tex]

Number 5:

  • [tex] \big[x - (\text-6)\big]^2 + \big[y - (\text-3)\big]^2 = (4\sqrt3\,)^2 [/tex]

  • [tex] (x + 6)^2 + (y + 3)^2 = 16(3) [/tex]

  • [tex] (x + 6)^2 + (y + 3)^2 = 48 [/tex]

Number 6:

  • [tex] (x - 0)^2 + (y -6)^2 = 9^2 [/tex]

  • [tex] x^2 + (y - 6)^2 = 81 [/tex]

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"Part B"

Number 7:

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

  • [tex] (x - 3)^2 + \big[y - (\text-4)\big]^2 = r^2 [/tex]

  • [tex] (x - 3)^2 + (y + 4)^2 = r^2 [/tex]

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

  • [tex] (6 - 3)^2 + (2 + 4)^2 = r^2 [/tex]

  • [tex] (3)^2 + (6)^2 = r^2 [/tex]

  • [tex] 9 + 36 = r^2 [/tex]

  • [tex] 45 = r^2 [/tex]

Substitute the square of the radius to the equation.

  • [tex] (x - 3)^2 + (y + 4)^2 = 45 [/tex]

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Number 8:

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

  • [tex] (x - 5)^2 + (y - 1)^2 = r^2 [/tex]

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

  • [tex] (8 - 5)^2 + (\text-2 - 1)^2 = r^2 [/tex]

  • [tex] (3)^2 + (\text-3)^2 = r^2 [/tex]

  • [tex] 9 + 9 = r^2 [/tex]

  • [tex] 18 = r^2 [/tex]

Substitute the square of the radius to the equation.

  • [tex] (x - 5)^2 + (y - 1)^2 = 18 [/tex]

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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.

  • [tex] (x - 0)^2 + (y - 0)^2 = r^2 [/tex]

  • [tex] x^2 + y^2 = r^2 [/tex]

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

  • [tex] 4^2 + 3^2 = r^2 [/tex]

  • [tex] 16 + 9 = r^2 [/tex]

  • [tex] 25 = r^2 [/tex]

Substitute the square of the radius to the equation.

  • [tex] x^2 + y^2 = 25 [/tex]

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Number 10:

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

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

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

  • [tex] \rm Center = \bigg(\frac{\,0\,}2,\, \frac{\,0\,}2 \bigg) \\ [/tex]

  • [tex] \rm Center = (0,0) [/tex]

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

  • [tex] (x - 0)^2 + (y - 0)^2 = r^2 [/tex]

  • [tex] x^2 + y^2 = r^2 [/tex]

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

  • [tex] 4^2 + 0^2 = r^2 [/tex]

  • [tex] 16 + 0 = r^2 [/tex]

  • [tex] 16 = r^2 [/tex]

Substitute the square of the radius to the equation.

  • [tex] x^2 + y^2 = 16 [/tex]

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