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10-1.Circle and System of Circles
hard
વર્તૂળો ${(x - 1)^2} + {(y - 3)^2} = {r^2}$ અને ${x^2} + {y^2} - 8x + 2y + 8 = 0$ બે ભિન્ન બિંદુમાં છેદે તો,
A
$2 < r < 8$
B
$2 < r < 8$
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
