The circles x2+y2=r2 and x2+y2−10x+16=0 intersect in real and distinct points and satisfies the condition a<r<b then (a,b) is given by
A
(2,10)
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B
(2,8)
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C
(0,4)
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D
none of these
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Solution
The correct option is D(2,8) C1(0,0) and r1=r C2(5,0) and r1=3 Since, circle intersect each other Then, |r1−r2|<C1C2<|r1+r2| ⇒|r−3|<5<|r+3| ⇒2<r<8 Ans: B