Question

Two circles touch internally.The sum of their areas is $116\pi c{m}^2$ and the distance between their centres is 6cm. Find the radii of the circles.

Answer 1

Sum of area of two circles $=116\times \pi c{m}^2$
Distance between the centers of two circles = 6 cm
Since the two circles are touching internally, the distance between the centers of the two circles =
(Radius of bigger circle) - (Radius of smaller circle)
Let r =radius of bigger circle
Then radius of small circle = r - 6
Area of big circle = 3.14 * r*r
and area of small circle $=116\times \left(\pi \right)$....(given)
$=\left(\pi \right)\times (r-6{)}^2=\left(\frac{\pi }{4}\right)\times (r^2-12r+36)$
Sum of area of two circles $=\left(\pi \right)\times r^2+\left[(\frac{\pi }{4}\times r^2-12r+36)\right]$
$=\left(\pi \right)\times (2r^2-12r+36)$
Dividing both side of equation by $2\times \left(\pi \right)$ we get
$r^2-6r-40=0$
by factorising left hand side of the equation we get:
(r - 10)(r + 4) = 0
Therefore r = 10, or r=-4
As radius cannot be negative, the radius of bigger circle = 10 cm
And the radius of smaller circle = 10 - 6 = 4 cm