Question

The circle $(x-2{)}^2+(y-5{)}^2=a^2$ will be inside the circle $(x-3{)}^2+(y-6{)}^2=b^2$ if

• A. $b>a+\sqrt{2}$
• B. $b>a-\sqrt{2}$
• C. $a<\sqrt{2}-b$
• D. $a+b>\sqrt{2}$

Answer 1

Answer: (A)

$b>a+\sqrt{2}$

The circle. $(x-2{)}^2+(y-5{)}^2=a^2$ will be inside $(x-3{)}^2+(y-6{)}^2=b^2$ if

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

Now, $c_1=(2,5),c_2=(3,6)$

$r_1=a,r_2=b$

so $l\left(c_1{c}_2\right)=\sqrt{(2-3{)}^2+(5-6{)}^2}=\sqrt{2}$

so, $\sqrt{2}<|a-b|$

$|a-b|>\sqrt{2}$

square both sides

Case (1) :

$a>b$.

So, $|a-b|=a-b$ implies $a-b>\sqrt{2}$

Case (2) $a<b$

So, $|a-b|=b-a$, which implies $b-a>\sqrt{2}$

$(b>a+\sqrt{2})$

So, answer is $a>b+\sqrt{2}$ if $a>b$

and $b>a+\sqrt{2}$ if $b>a$.

So from given options answer is Option [A], that is $b>a+\sqrt{2}$