26. If the circle x² + y² + 4x + 22y + c = 0 bisects the circumference of the circle x² + y² – 2x + 8y – d = 0, then c + d is equal to

Question

26. If the circle x² + y² + 4x + 22y + c = 0 bisects the circumference of the circle
x² + y² – 2x + 8y – d = 0, then c + d is equal to :
(a) 60
(b) 50
(c) 40
(d) 30

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Quinn 4 months 2022-01-06T02:15:33+00:00 2 Answers 0 views 0

Answers ( )

    0
    2022-01-06T02:16:36+00:00

    Answer

    => c + d = 50

    Given that, circle x² + y² + 4x + 22y + c = 0 bisects the circumference of the circle

    x² + y² – 2x + 8y – d = 0

    The common chord of the given circle is

    S1 – S2 =0

    => x² + y² + 4x + 22y + c – x² – y² + 2x – 8y + d = 0

    => 6x + 14y + c + d = 0. ….(1)

    so, Eq. (1) passes through the centre of the second circle i.e., (1, -4)

    6 – 56 + c + d = 0

    => c + d = 50

    Note: If S1 and S2 are the equation of the two circle, then equation of common chord is S1 – S2 = 0.

    0
    2022-01-06T02:17:20+00:00

    \huge{\boxed{\overline{\underline{\mathfrak{\fcolorbox{cyan}{blue}{Answer}}}}}}

    => c + d = 50

    Given that, circle x² + y² + 4x + 22y + c = 0 bisects the circumference of the circle

    x² + y² – 2x + 8y – d = 0

    The common chord of the given circle is

    S1 – S2 =0

    => x² + y² + 4x + 22y + c – x² – y² + 2x – 8y + d = 0

    => 6x + 14y + c + d = 0. ….(1)

    so, Eq. (1) passes through the centre of the second circle i.e., (1, -4)

    6 – 56 + c + d = 0

    => c + d = 50

    Note: If S1 and S2 are the equation of the two circle, then equation of common chord is S1 – S2 = 0.

    silentlover45.❤️

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