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Let a circle C of radius 5 lie below the x-axis. The line L1 : 4x + 3y + 2 = 0 passes through the centre P of the circle C and intersects the line L2 : 3x – 4y – 11 = 0 at Q. The line L2 touches C at the point Q. Then the distance of P from the line 5x – 12y + 51 = 0 is……….

0 3 Views | Posted 2 months ago
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    Answered by

    Payal Gupta | Contributor-Level 10

    2 months ago

    L 1 : 4 x + 3 y + 2 = 0  

    L 2 : 3 x − 4 y − 1 1 = 0               

    Since circle C touches the line L2 at Q intersection point L1 and L2 is (1, -2)

    P lies of L1

    ∴ P ( x , − 1 3 ( 2 + 4 x ) )               

    Now,

    PQ = 5 ? (x – 1)2 + ( 4 x + 2 3 − 2 ) 2 = 2 5  

    ⇒ x = 4 , − 2                     

    ? The circle lies below the axis

    y = -6

    p (4, -6)

    Now distance of P from 5x – 12 y + 51 = 0

    = | 2 0 + 7 2 + 5 1 1 3 | = 1 4 3 1 3 = 1 1                

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