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Question: If two circles \(( x - 1 ) ^ { 2 } + ( y - 3 ) ^ { 2 } = r ^ { 2 }\) and \(x ^ { 2 } + y ^ { 2 } - 8...

If two circles (x−1)2+(y−3)2=r2( x - 1 ) ^ { 2 } + ( y - 3 ) ^ { 2 } = r ^ { 2 } and x2+y2−8x+2y+8=0x ^ { 2 } + y ^ { 2 } - 8 x + 2 y + 8 = 0 intersect in two distinct points, then

A

2<r<82 < r < 8

B

r=2r = 2

C

r<2r < 2

D

r>2r > 2

Answer

2<r<82 < r < 8

Explanation

Solution

When two circles intersect each other, then difference between their radii < Distance between their centres

⇒r−3<5\Rightarrow \quad r - 3 < 5

⇒r<8\Rightarrow \quad r < 8 …..(i)

Sum of their radii > Distance between their centres

⇒r+3>5\Rightarrow \quad r + 3 > 5

⇒r>2\Rightarrow \quad r > 2 …..(ii)

Hence by (i) and (ii), 2 < r < 8.