The equation of a circle whose center is at (4, 0) and radius is length 2√(3) is (x – 4)² + y² = 2√3 (x – 4)² + y² = 12 (x + 4)² + y² = 12

Question

The equation of a circle whose center is at (4, 0) and radius is length 2√(3) is (x – 4)² + y² = 2√3 (x – 4)² + y² = 12 (x + 4)² + y² = 12

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    0
    2021-09-04T18:34:11+00:00

    The equation of a circle is (x-a)²+(y-b)²=R²
    (x-4)²+(y-0)²=(2√3)²
    (x-4)²+y²=12

    0
    2021-09-04T18:34:45+00:00

    Answer:

    Equation of the circle is

    (x-4)^2+y^2=12

    Step-by-step explanation:

    We have been given that

    center of the circle = (4,0)

    Radius of the circle = 2\sqrt3

    The standard form of a circle is given by

    (x-h)^2+(y-k)^2=r^2

    Here, (h,k) is the center and r is the radius. Thus, we have

    h = 4, k = 0, r = 2√(3)

    Substituting these values in the above equation, we get

    (x-4)^2+(y-0)^2=(2\sqrt3)^2

    Simplifying, we get

    (x-4)^2+y^2=12

    Therefore, equation of the circle is

    (x-4)^2+y^2=12

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