1 1 vote Two circles, each of radius $4$ cm, touch externally. Each of these two circles is touched externally by a third circle. If these three circles have a common tangent, then the radius of the third circle, in cm, is $\sqrt{2}$ $\frac{\pi }{3}$ $\frac{1}{\sqrt{2}}$ $1$ Quantitative Aptitude cat2019-2 quantitative-aptitude geometry + – go_editor 14.2k points 1.8k views answer comment Share Follow Print 0 reply Please log in or register to add a comment.
1 1 vote Let $r$ be the radius of the third circle. Then, From the above diagram, we get $AE= GC + CD$ $\Rightarrow 4 = r + CD $ $ \Rightarrow CD = ( 4-r )$ In $ \triangle ADC,$ $ \angle D = 90^{\circ},$ using Pythagoras theorem. $ \boxed{(AC)^{2} = (CD)^{2} + (AD)^{2}} $ $(4+r)^{2} = (4-r)^{2} + (4)^{2}$ $ \Rightarrow 16 + r^{2} + 8r = 16 + r^{2} – 8r + 16$ $ \Rightarrow 8r = -8r + 16 $ $\Rightarrow 8r + 8r = 16$ $ \Rightarrow 16r = 16$ $ \Rightarrow r = 1$ $ \therefore $ The radius of the third circle is $1\;\text{cm}.$ Correct Answer: D Anjana5051 answered Jul 31, 2021 • edited Aug 25, 2021 by Lakshman Bhaiya Anjana5051 12.1k points comment Share Follow 0 reply Please log in or register to add a comment.