If the circles x2+y2−16x−20y+164=r2 and (x−4)2+(y−7)2=36 intersect at two distinct points, then

If the circles x2+y216x20y+164=r2 and (x4)2+(y7)2=36 intersect at two distinct points, then

  1. A

    1<r<11

  2. B

    r>11

  3. C

    0<r<1

  4. D

    r=11

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    Solution:

    Given equation of circle is x2+y216x20y+164=r2(x8)2+(y10)2=r2

    C1=(8,10) and R1=r

    for the second circle C2=(4,7),R2=6

    now, C1C2=(84)2+(107)2=25=5

    Since the circles intersect at two distinct points So, 

    R1+R2>C1C2>R1R2

    r+6>5>|r6|1<r<11

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