Question

Two circles touch each other externally.The sum of their areas is 58pi cm2 and the distance between their centres is 10 cm. Find the radii of the two circles.

Answer 1

ANSWER:

r = 3 cm, 7 cm

STEP-BY-STEP EXPLANATION:

Given that two circles touch each other externally. The sum of their areas is 58pi cm² and the distance between their centres is 10 cm.

We need to find out the radii of the two circles.

Let's say the radius of the circle is r.

As the distance between their centres is 10 cm, means the radius of the other circle is (10 - r).

Area of circle is πr² and area of other circle is π(10 - r)². And the sum of the area is 58π cm².

Now,

Area of circle + Area of other circle = 58π

πr² + π(10 - r)² = 58π

πr² + π(100 + r² - 20) = 58π

π(r² + 100 + r² - 20r) = 58π

2r² - 20r + 100 = 58

2r² - 20r + 42 = 0

r² - 10r + 21 = 0

Spilt the middle term in such a way that it's sum is -10 and product is 21.

r² - 7r - 3r + 21 = 0

r(r - 7) -3(r - 7) = 0

(r - 7)(r - 3) = 0

r = 7, 3

Therefore, the radius of the one circle is 3 cm and other circle is 7 cm.

Answer 2

Given,

Two circles touch each other externally. (i.e., two circles are placed side by side touching each other and their centres are in a line).

The sum of areas of two circles is $58\pi cm^{2}$.

The distance between centres of two circles is $10cm$.

Let the radius of one circle be $'x'cm$.

We know that, formula for area of a circle $=\pi r^{2}$.

Area of first circle $=\pi x^{2}$.

Then, the radius of circle will be $'10-x'cm$.

Area of second circle $=\pi (10-x)^{2}$.

According to given information:

$\pi x^{2} +\pi (10-x)^{2} =58\pi$

$\pi [x^{2} +(10-x)^{2}]=58\pi$

$x^{2} +(10-x)^{2} =58$ (by dividing $\pi$ on both sides).

$x^{2} +[10^{2} +x^{2} -(2\times10\times x)]=58$

$x^{2} +100+x^{2} -20x=58$

$2x^{2} -20x+100-58=0$

$2x^{2} -20x+42=0$

$2(x^{2} -10x+21)=0$

$(x^{2} -10x+21)=(0/2)$

$x^{2} -7x-3x+21=0$

$x(x-7)-3(x-7)=0$

$(x-3)(x-7)=0$

$(x-3=0) ; (x-7=0)$

$(x=3) ; (x=7)$

So, the radius of one circle $=3cm$.

and the radius of the other circle $=7cm$.