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Question

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A) True B) False

Answer
Verified

Given, the length of common tangent to the circles = $15$cm and the radii of the two circles are $12$cm and $4$cm. We have to find the distance between their centers. Let A and B be the centers of the two circles respectively and CD be the common tangent to the circles. Let us draw EB $\parallel $CD to make rectangle BDCE where CD=EB=$15$cm and EC=BD= $4$cm

Now from the figure, AE=AC-CE

$ \Rightarrow {\text{AE = 12 - 4 = 8}}$ cm

Now In the triangle BEA, We know EB and AE but we have to find AB.

Let AB=x cm. Since ∆ BEA is a right-angled triangle, we can use the Pythagoras theorem to find AB. According to the Pythagoras theorem, “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides.”It can be written as-

$ \Rightarrow $ \[{{\text{H}}^2} = {{\text{P}}^2}{\text{ + }}{{\text{B}}^2}\]

Where H is the hypotenuse, P is the perpendicular and B is the base of the triangle.

In triangle BEA, H=AB, P=EB and B=AE

Then on using the formula, we get

$ \Rightarrow $ ${\left( {{\text{AE}}} \right)^2} + {\left( {{\text{EB}}} \right)^2} = {\text{A}}{{\text{B}}^2}$

$

\Rightarrow {{\text{x}}^2} = {\left( 8 \right)^2} + {\left( {15} \right)^2} = 64 + 225 = 289 \\

\Rightarrow {\text{x}} = \sqrt {289} = 17 \\

$

AB=$17$ cm= the distance between the centers of the two circles. So the statement is true.

Hence the correct answer is ‘A’.

$ \Rightarrow $ d=$\sqrt {{{15}^2} + {{\left( {12 - 4} \right)}^2}} = \sqrt {225 + {8^2}} = \sqrt {225 + 64} $

$ \Rightarrow $ d=$\sqrt {289} = 17$ cm

But If the circles have transverse common tangent (see the figure)-

Then we use formula for distance between their centers ${\text{d = }}\sqrt {{{\text{t}}^2} + {{\left( {{{\text{r}}_1}{\text{ + }}{{\text{r}}_2}} \right)}^2}} $ where the notations have same meaning.

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