A is the centre of a circle whose radius is 8 | vedclass

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(A) The radius of the first circle centered at $A$ is $r_1 = 8$.The diameter of the second circle centered at $B$ is $8$,so its radius is $r_2 = \frac

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(A) The radius of the first circle centered at $A$ is $r_1 = 8$.The diameter of the second circle centered at $B$ is $8$,so its radius is $r_2 = \frac{8}{2} = 4$.Since the two circles touch externally,the distance between their centers $A$ and $B$ is the sum of their radii:$AB = r_1 + r_2 = 8 + 4 = 12$.Now,we need to find the area of a circle with diameter $AB = 12$.The radius of this new circle is $R = \frac{AB}{2} = \frac{12}{2} = 6$.The area of the circle is given by the formula $\pi R^2$.Area $= \pi 	imes (6)^2 = 36\pi$.

$A$ is the centre of a circle whose radius is $8$ and $B$ is the centre of a circle whose diameter is $8$. If these two circles touch externally,then the area of the circle with diameter $AB$ is $.........\pi$.

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