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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
(2019)
  • a)
    √2
  • b)
    π/3
  • c)
    1/√2
  • d)
    1
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
Two circles, each of radius 4 cm, touch externally. Each of these two ...

From the figure, OC = AD – r = 4 – r
Applying Pythagoras theorem in right triangle AOC, we get
AC2 = AO2 + OC2
⇒ (4 + r)2 = 42 + (4 – r)2
⇒ (4 + r)2 – (4 – r)2 = 16
⇒ 8 × 2r = 16
∴ r = 1
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Most Upvoted Answer
Two circles, each of radius 4 cm, touch externally. Each of these two ...
Let the centers of the two smaller circles be $A$ and $B$, and the center of the larger circle be $C$. Let the tangency points of the two smaller circles with the larger circle be $D$ and $E$. Let the point of tangency of the common tangent with circle $A$ be $F$ and the point of tangency of the common tangent with circle $B$ be $G$. Let the point of tangency of the common tangent with the larger circle be $H$. We can draw the following diagram:
[asy] draw((-4,0)--(4,0),linewidth(1)); draw((-2,0)--(-2,2*sqrt(3)),linewidth(1)); draw((2,0)--(2,2*sqrt(3)),linewidth(1)); draw((-2,2*sqrt(3))--(2,2*sqrt(3)),linewidth(1)); draw((-2,0)--(0,4),linewidth(1)); draw((2,0)--(0,4),linewidth(1)); draw((0,4)--(0,2*sqrt(3)),linewidth(1)); label("$A$",(-2,0),SW); label("$B$",(2,0),SE); label("$C$",(0,2*sqrt(3)),N); label("$D$",(0,4),N); label("$E$",(-2,2*sqrt(3)),W); label("$F$",(-1,2),W); label("$G$",(1,2),E); label("$H$",(0,4),N); dot((-2,0)); dot((2,0)); dot((0,2*sqrt(3))); dot((0,4)); dot((-2,2*sqrt(3))); dot((-1,2)); dot((1,2)); [/asy]
Since $\overline{AF}$ is tangent to circle $A$, $\angle AFE=\angle AEF=90^\circ$. Similarly, $\angle BEG=\angle BGE=90^\circ$. Since $\overline{AF}$ and $\overline{BG}$ are parallel, $\angle EAF=\angle EBG$. Thus, $\triangle AEF\sim\triangle BEG$. By AAA, $\triangle AEF\sim\triangle BEG$. Since $\triangle AEF\sim\triangle CDH$ and $\triangle BEG\sim\triangle CDH$, we have
$$\frac{EF}{GH}=\frac{AF}{BG}=\frac{AE}{BE}=\frac{CD}{CD} = 1.$$Since $EF=GH$, we have $EF=GH=4$. Let the radius of the larger circle be $r$. By the Pythagorean Theorem on right $\triangle ACF$, we have
$$r^2=AC^2=AF^2+CF^2=(4+r)^2+4^2=2r^2+16r+32.$$Solving for $r$, we see that $r=\boxed{016}$.
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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(2019)a)√2b)π/3c)1/√2d)1Correct answer is option 'D'. Can you explain this answer?
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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(2019)a)√2b)π/3c)1/√2d)1Correct answer is option 'D'. Can you explain this answer? for CAT 2024 is part of CAT preparation. The Question and answers have been prepared according to the CAT exam syllabus. Information about 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(2019)a)√2b)π/3c)1/√2d)1Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for CAT 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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(2019)a)√2b)π/3c)1/√2d)1Correct answer is option 'D'. Can you explain this answer?.
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