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Question

# Let c1& c2 be the centers and r1& r2 be the radius of two circles. Then Cases Conditions p. 1.|r1âˆ’r2|< c1c2< r1+r2 q. 2. |r1âˆ’r2|=c1c2 r. 3.c1c2< |r1+r2| s. 4. r1+r2< c1r2 t. 5. r1+r2=c1r2

A

p - 5, q - 4 , r - 1, s - 2 , t - 3

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B

p - 4, q - 5 , r - 1, s - 3 , t - 2

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C

p - 1, q - 2 , r - 3, s - 4 , t - 5

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D

p - 4, q - 5 , r - 1 , s - 2 , t - 3

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Solution

## The correct option is D p - 4, q - 5 , r - 1 , s - 2 , t - 3 Two circles with centers c1(g1,f1) and c2(g2,f2) and radiir1 andr2 respectively, then we have following results. p. Here, two circles do not interact with each other and distance between the centers is greater than the sum of the radii. So,c1c2> r1+r2 (p-4) q. Here,two circles touch each other externally. i.e., distance between centers is equal to the sum of the radii. c1c2=r1+r2 (q-5) r. Here, circle intersects at two real and distinct points, i.e., the distance between the centers is less than the sum of radii c1c2< r1+r2 (r-1) s. Here, circlrs touch each other internally; distance between centers is equal to the difference of their radii. c1c2=|r1r2| { r1≠ r2} t. Herw, one circles with smaller radius lies inside the other i.e., distance between the centers is less than the difference of the radii. c1c2< |r1−r2|{ r1≠ r2} (t-3) So, correct matched is p−4,q−5,r−1,s−2,t−3

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