Circles {x^2} + {y^2} - 10x + 16 = 0 and {x^2} + {y^2} = {r^2} intersect each other in two distinct

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10-1.Circle and System of Circles

hard

The circles ${x^2} + {y^2} - 10x + 16 = 0$ and ${x^2} + {y^2} = {r^2}$ intersect each other in two distinct points, if

A

$r < 2$

B

$r > 8$

C

$2 < r < 8$

D

$2 \le r \le 8$

(IIT-1994)

Solution

(c) If $d$ is the distance between the centres of two circles of radii ${r_1}$ and ${r_26}$, then they intersect in two distinct points, if

$| {r_1} – {r_2}|\; < d < ({r_1} + {r_2})$

Here, ${r_1} = \sqrt {25 – 16} = 3$

${r_2} = r \Rightarrow \;|3 – r|\; < 5 < \;|3 + r|\; \Rightarrow 2 < r < 8$.

Standard 11

Mathematics

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