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Q.

Circles of radii 36 and 9 touch externally. The radius of the circle which touches the two circles externally and also their common tangent is :

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a

2

b

4

c

\sqrt{17}

d

\sqrt{18}

answer is B.

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Detailed Solution

Question Image

Let the center of the small circle with radius 9 as A and center of the large circle radius 36 as C. B and D are the points of contact with their common tangent.
AB=9,CD=36.
Let the center of the circle which touches both the small and large circle as O and the point of contact of this circle on the tangent is G. So its radius is OG(r).
We produce the line segment MN which passes through O and parallel and equal to BD. We also produce the line segment AL which is parallel and equal to BD. We know that the radius of a circle is perpendicular to its tangent. So BD as well as its parallel lines MN and AL will be perpendicular to both radii AB and CD.
Let BG = x, BD = d

\Rightarrow

GD = d−x

\Rightarrow

BD = AL = MN = d, OM = d−x, ON = x.
AC = AB+CD = 36+9 = 45
CL = CD−LD = CD−AB = 36−9 = 27.
In the right-angled triangle ALC,

{AC}^{2} = {AL}^{2} + {LC}^{2}

\Rightarrow 45^{2} = d^{2} - 27^{2}

\Rightarrow d = 36

In the right-angled triangle AON, AO = 9 + OG = 9 + r and AN = AB − NB = AB − OG = 9−r.

{AO}^{2} = {AN}^{2} + {NO}^{2}

\Rightarrow {(9 + r)}^{2} = {(9 - r)}^{2} + x^{2}

\Rightarrow x = 6 \sqrt{r}

In the right-angled triangle COM, CO=CD+OG=36+r and CM=CD−DM=CD−OG=36−r.

{CO}^{2} = {CM}^{2} + {MO}^{2}

\Rightarrow {(36 + r)}^{2} = {(36 - r)}^{2} + {(d - x)}^{2}

\Rightarrow d - x = 12 \sqrt{r}

\Rightarrow 36 - 6 \sqrt{r} = 12 \sqrt{r}

\Rightarrow \sqrt{r} = 2

\Rightarrow r = 4

So, Option 2 is correct.

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Circles of radii 36 and 9 touch externally. The radius of the circle which touches the two circles externally and also their common tangent is :