CAT 2019 Slot 2QA Question 7

2 CirclesEasy

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

Answer & solution

Correct answer: 1

  • A

    π/3

  • 1

  • C

    1/√2

  • D

    √2

Solution

Easy

All three circles sit on the same straight tangent line, so each centre is at a height equal to its radius. Connect the centre of one big circle (radius 44) to the small circle (radius xx); the centre-to-centre distance is 4+x4+x (external touch) and the horizontal/vertical legs come from the radii. Apply the Pythagorean theorem.

4 x
1

Set up the right triangle. Let the small circle have radius xx. Each centre lies above the common tangent at a height equal to its radius. Taking a big-circle centre AA (height 44) and the small-circle centre CC (height xx), the right triangle on the line has:

hypotenuse AC=4+x(circles touch externally)vertical leg=4x(difference of heights)\begin{aligned} &\text{hypotenuse } AC=4+x \quad\text{(circles touch externally)}\\ &\text{vertical leg}=4-x \quad\text{(difference of heights)} \end{aligned}
2

Find the horizontal leg. The two equal circles touch externally, so the foot of the big circle on the tangent is 44 from the contact point — the horizontal distance from AA to CC works out to 44. Apply the Pythagorean theorem in ADC\triangle ADC.

(4+x)2=42+(4x)2\begin{aligned} &(4+x)^{2}=4^{2}+(4-x)^{2} \end{aligned}
3

Solve for xx. Expand both squares from step 2.

16+8x+x2=16+168x+x2 8x=168x 16x=16  x=1\begin{aligned} &16+8x+x^{2}=16+16-8x+x^{2}\\ &\Rightarrow\ 8x=16-8x\\ &\Rightarrow\ 16x=16\ \Rightarrow\ x=1 \end{aligned}
Radius of the third circle=1 cm\text{Radius of the third circle}=1\ \text{cm}

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CAT 2019 Slot 2 QA Q7: Two circles, each of radius 4 cm, touch externally. Each of these two circles is touched externally by a third — Solution | TheCATExam