Two circles touch externally- The sum of their areas is 130- sq- cm- and the distance between their centres is 14cm- Find the radii of the circles-

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Theorem on Touching Circles

Two circles t...

Question

Two circles touch externally. The sum of their areas is $130 π$ sq. cm. and the distance between their centres is $14 c m$ . Find the radii of the circles.

11

cm and

3

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1

cm and

3

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11

cm and

13

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None of these

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Solution

The correct option is A $11$ cm and $3$ cm
If two circles touch externally, then the distance between their centers is equal to the sum of their radii. Let the radii of the two circles be $r_{1}$ cm and $r_{2}$ cm respectively.

Let $C_{1}$ and $C_{2}$ be the centres of the given circles. Then,
$C_{1} C_{2} = r_{1} + r_{2}$ [∵ $C_{1} C_{2} = 14$ cm (given)]
⇒ $14 = r_{1} + r_{2}$
⇒ $r_{1} + r_{2} = 14$ ....(i)

It is given that the sum of the areas of two circles is equal to $130 π c m^{2}$ .
∴ $π (r_{1})^{2} + π (r_{2})^{2} = 130 π$
⇒ $(r_{2})^{2} + (r_{2}) 2 = 130$ ...(ii)

Now, $(r_{1} + r_{2})^{2} = (r_{1})^{2} + (r_{2})^{2} + 2 r_{1} r_{2}$
⇒ $142 = 130 + 2 r_{1} r_{2}$ [Using (i) and (ii)]
⇒ $196 - 130 = 2 r_{1} r_{2}$
⇒ $r_{1} r_{2} = 33$ ....(iii)

Now,
$(r_{1} - r_{2})^{2} = (r_{1})^{2} + (r_{2})^{2} - 2 r_{1} r_{2}$
⇒ $(r_{1} - r_{2})^{2} = 130 - 2 \times 33$ [Using (ii) and (iii)]
⇒ $(r_{1} - r_{2})^{2} = 64$
⇒ $r_{1} - r_{2} = 8$ ....(iv)

Solving (i) and (iv), we get $r_{1} = 11$ cm and $r_{2} = 3$ cm. Hence, the radii of the two circles are $11$ cm and $3$ cm.

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Two circles touch externally. The sum of their areas is 130π sq. cm. and the distance between their centres is 14cm. Find the radii of the circles.

Two circles touch externally. The sum of their areas is $130 π$ sq. cm. and the distance between their centres is $14 c m$ . Find the radii of the circles.