Q.

The equation of the circle whose radius is 5 and which touches the circle x2+y2−2x−4y−20=0 at the point (5,5) is

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a

x2+y2+18x+16y+120=0

b

x2+y2−18x−16y+120=0

c

x2+y2−18x+16y+120=0

d

x2+y2+18x−16y+120=0

answer is B.

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Detailed Solution

Centre of the given circle is (1,2) and its radius =1+4+20=5 Since the radii of the two circles are  cqual, therefore these will touch externally and the point of contact will lie mid - way between the two  centres. If (h,k) is the centre of the circle, thenh+12=5,k+22=5 ∴h=9,k=8∴ its equation is x-92+y-82=52 i.e., x2+y2−18x−16y+120=0
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