On the same side of a straight line three circles are drawn as follows:
a circle with a radius of $4$ cm is tangent to the line, the other two circles are equal, and each is tangent to the line and to the other two circles. The radius of the equal circles is:

24

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Solution

The correct option is D $16$
Let radius of two equal circle be 'r'
Radius of small circle $= 4 ㎝$
In $△ O^{'} M P$
$O^{'} N = r$
$O^{'} M = O N - M N = r - 4$
$O^{'} P = r + 4$
Using Pythagoras theorem, we get
${(r - 4)}^{2} + r^{2} = {(r + 4)}^{2} \Rightarrow r^{2} + 16 - 8 r + r^{2} = r^{2} + 16 - 8 r$
$\Rightarrow r^{2} - 16 r = 0 \Rightarrow r (r - 16) = 0 \Rightarrow r = 0, 16$
$∴$ Radii of equal circles $= 16 ㎝$

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On the same side of a straight line three circles are drawn as follows:a circle with a radius of 4 cm is tangent to the line, the other two circles are equal, and each is tangent to the line and to the other two circles. The radius of the equal circles is:

The correct option is D 16Let radius of two equal circle be 'r'Radius of small circle=4㎝In △O′MPO′N=rO′M=ON−MN=r−4O′P=r+4Using Pythagoras theorem, we get(r−4)2+r2=(r+4)2⇒r2+16−8r+r2=r2+16−8r⇒r2−16r=0⇒r(r−16)=0⇒r=0,16∴ Radii of equal circles=16㎝

On the same side of a straight line three circles are drawn as follows:
a circle with a radius of $4$ cm is tangent to the line, the other two circles are equal, and each is tangent to the line and to the other two circles. The radius of the equal circles is: