Question

Let $c_1\&c_2$ be the centers and $r_1\&r_2$ be the radius of two circles. Then

Cases Conditions

p. 1.$|r_1-r_2|<c_1{c}_2<r_1+r_2$

q. 2. $|r_1-r_2|=c_1{c}_2$

r. 3.$c_1{c}_2<|r_1+r_2|$

s.

4. $r_1+r_2<c_1{r}_2$

t. 5. $r_1+r_2=c_1{r}_2$

• A. p - 5, q - 4 , r - 1, s - 2 , t - 3
• B. p - 4, q - 5 , r - 1, s - 3 , t - 2
• C. p - 1, q - 2 , r - 3, s - 4 , t - 5
• D. p - 4, q - 5 , r - 1 , s - 2 , t - 3

Answer 1

The correct option is D

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

Two circles with centers $c_1(g_1,f_1)and{c}_2(g_2,f_2)andradii{r}_{1}and{r}_2$ 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,$c_1{c}_2>r_1+r_2$

(p-4)

q.

Here,two circles touch each other externally. i.e., distance between centers is equal to the sum of the

radii.

$c_1{c}_2=r_1+r_2$

(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

$c_1{c}_2<r_1+r_2$

(r-1)

s.

Here, circlrs touch each other internally; distance between centers is equal to the difference of their

radii.

$c_1{c}_2=|r_1{r}_2|$ $\{r_1\ne r_2\}$

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.

$c_1{c}_2<|r_1-r_2|\{r_1\ne r_2\}$

(t-3)

So, correct matched is

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