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