Identify the equation of the circle that has its center at (-27, 120) and passes through the origin.

Question

Identify the equation of the circle that has its center at (-27, 120) and passes through the origin.

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Athena 2 months 2021-10-05T12:38:55+00:00 1 Answer 0 views 0

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    2021-10-05T12:40:27+00:00

    The equation of the circle that has its center at (-27, 120) and passes through the origin is:

    (x + 27)^2 + (y - 120)^2 = 15129

    Solution:

    The equation of a circle is given as:

    (x-a)^2+(y-b)^2=r^2

    Where,

    (a, b) is the centre of the circle

    r is the radius

    We have the centre of the circle (-27, 120)

    Therefore,

    a = -27

    b = 120

    Given that, it passes through origin. which means, (x, y) = (0, 0)

    Substitute (a, b) = (-27, 120) and (x, y) = (0, 0) in eqn

    (0 + 27)^2 + (0 - 120)^2 = r^2\\\\729 + 14400 = r^2\\\\r^2 = 15129

    Substitute r^2 = 15129 and (a, b) = (-27, 120) in eqn

    (x + 27)^2 + (y - 120)^2 = 15129

    Thus the equation of circle is found

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