If two circles {(x - 1)^2} + {(y - 3)^2} = {r^2} and {x^2} + {y^2} - 8x + 2y + 8 = 0 intersect in tw

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10-1.Circle and System of Circles

hard

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

A

$2 < r < 8$

B

$r = 2$

C

$r < 2$

D

$r > 2$

(AIEEE-2003) (IIT-1989)

Solution

(a) When two circles intersect each other, then

Difference between their radii $r – 3 < 5$ $⇒$ $r = 8$…..$(i)$

Difference between their radiiSum of their radii > Distance between centres ….$(ii)$

$ \Rightarrow $ $r + 3 > 5$

$\Rightarrow r > 2$

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

Standard 11

Mathematics

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