Question

Two circles touch each other externally. The sum of their areas is $130\pi {\text{cm}}^2$ and the distance between their centres is $14$ cm. Find the diameters of the circle.

• A. $22,8$
• B. $22,6$
• C. $26,8$
• D. $22,4$

Answer 1

The correct option is B $22,6$

Let the radius of the two circles are $a$ and $b$ cm

The circles touch each other externally.

So, the distance between their centers $=a+b$

$a+b=14$ ................... (1)

Sum of their areas $=\pi a^2+\pi b^2$

$\pi a^2+\pi b^2=130\pi$

$a^2+b^2=130$ .................... (2)

From equation (1),

$b=14-a$ ............. (3)

From (2) and (3),

$a^2+(14-a{)}^2=130$

$a^2+a^2+196-28a=130$

$2a^2+66-28a=0$

$a^2+33-14a=0$

$a^2-11a-3a+33=0$

$(a-11)(a-3)=0$

$a=3,11$

If $a=3$, then $b=14-3=11$

If $a=11$, then $b=14-11=3$

So, the radius of the two circles are $3$ and $11$ cm

And their diameters are $6$ and $22$ cm.