Give the standard form of the circle whose center C and radius r are given 1. C (0,0) r=10 2. C(2,6) r=9 3. C(-7,2) r=

Question

Give the standard form of the circle whose center C and radius r are given

1. C (0,0) r=10
2. C(2,6) r=9
3. C(-7,2) r=15​

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Mackenzie 2022-07-06T05:36:24+00:00 1 Answer 0

Answers ( 1 )

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    2022-07-06T05:38:11+00:00

    ✒️CIRCLE EQUATIONS

    [tex]••••••••••••••••••••••••••••••••••••••••••••••••••[/tex]
    [tex] \large\underline{\mathbb{ANSWER}:} [/tex]
    [tex] \qquad \large \rm 1) \; x^2 + y^2 = 100 [/tex]
    [tex] \qquad \large \rm 2) \; (x-2)^2 + (y-6)^2 = 81 [/tex]
    [tex] \qquad \large \rm 3) \; (x+7)^2 + (y-2)^2 = 225 [/tex]
    *Please read and understand my solution. Don’t just rely on my direct answer*
    [tex]••••••••••••••••••••••••••••••••••••••••••••••••••[/tex]
    [tex] \large\underline{\mathbb{SOLUTION}:} [/tex]
    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 the equation.
    [tex]••••••••••••••••••••••••••••••••••••••••••••••••••[/tex]

    Number 1:

    Substitute (h,k) as (0,0) and r as 10.
    • [tex] (x-0)^2 + (y-0)^2 = 10^2 [/tex]
    • [tex] x^2 + y^2 = 100 [/tex]
    Therefore, the given equation in standard form is + = 100
    [tex]••••••••••••••••••••••••••••••••••••••••••••••••••[/tex]

    Number 2:

    Substitute (h,k) as (2,6) and r as 9.
    • [tex] (x-2)^2 + (y-6)^2 = 9^2 [/tex]
    • [tex] (x-2)^2 + (y-6)^2 = 81 [/tex]
    Therefore, the given equation in standard form is (x-2)² + (y-6)² = 81
    [tex]••••••••••••••••••••••••••••••••••••••••••••••••••[/tex]

    Number 3:

    Substitute (h,k) as (-7,2) and r as 15.
    • [tex] \big[x-(\text-7)\big]^2 + (y-2)^2 = 15^2 [/tex]
    • [tex] (x+7)^2 + (y-2)^2 = 225 [/tex]
    Therefore, the given equation in standard form is (x+7)² + (y-2)² = 225
    [tex]••••••••••••••••••••••••••••••••••••••••••••••••••[/tex]
    (ノ^_^)ノ [tex] \large\qquad\qquad\qquad\tt 2/24 /2022 [/tex]

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