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The condition that the circles which pas...

The condition that the circles which passes through the points `(0,a),(0,-a)` and touch the line `y= mx+c` will cut orthogonally is

A

`c^(2) =a^(2) (1+m^(2))`

B

` c^(2) =a^(2) (2+ m ^(2))`

C

` c^(2) = a^(2) (3+m^(2))`

D

`c^(2) =a^(2) (4+ m^(2))`

Text Solution

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The correct Answer is:
B
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Knowledge Check

  • The condition that the two circles which passes through the points (0,a),(0,-a) and touch the line y = mx+ c will cut orthogonally is

    A
    `c^(2) = a^(2) (1 + m^(2)) `
    B
    `c^(2) = a^(2) (2 + m^(2))`
    C
    `c^(2)= a^(2) (3 + m^(2))`
    D
    `c^(2) = a^(2) (4 + m^(2))`

  • The equation of the circle which passes through the origin has its centre on the line x+y=4 and cuts orthogonally the circle x^2+y^2-4x+2y+4=0

    A
    `x^2+y^2-4x-4y=0`
    B
    `x^2+y^2-2x-6y=0`
    C
    `x^2+y^2-6x-2y=0`
    D
    `x^2+y^2+4x-12y=0`

  • The centre of the circle passing through the point (0, 1) and touching the curve y=x^(2) at (2, 4) is

    A
    `((16)/(5), (53)/(10))`
    B
    `((-2)/(3), (-4)/(3))`
    C
    `((-4)/(3), (2)/(3))`
    D
    `((-16)/(5), (53)/(10))`

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