Answer
423.9k+ views
Hint: We can solve the problem by using pythagoras theorem by making figure on given conditions or we can also solve using the formula ${\text{d = }}\sqrt {{{\text{t}}^2} + {{\left( {{{\text{r}}_1} - {{\text{r}}_2}} \right)}^2}} $ where d is the distance between the centers of two circles with radii ${{\text{r}}_1}$ and ${{\text{r}}_2}$, and t is the length of the common tangent. On putting the given values, you’ll get the answer.
Complete step by step answer:
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’.
Note:: We can also use the formula ${\text{d = }}\sqrt {{{\text{t}}^2} + {{\left( {{{\text{r}}_1} - {{\text{r}}_2}} \right)}^2}} $ where d is the distance between the centers of two circles with radii ${{\text{r}}_1}$ and${{\text{r}}_2}$, and t is the length of the common tangent and we’ll get the same answer. This formula is used when the circles have a direct common tangent. On putting the given values we get,
$ \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.
Complete step by step answer:
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
![seo images](https://www.vedantu.com/question-sets/5468f594-dd45-47eb-a8d1-17c08c2d04b96808806468609985702.png)
Now from the figure, AE=AC-CE
$ \Rightarrow {\text{AE = 12 - 4 = 8}}$ cm
![seo images](https://www.vedantu.com/question-sets/9c7f79f6-648d-447c-b7e7-9552a602a84b8665954775872648867.png)
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’.
Note:: We can also use the formula ${\text{d = }}\sqrt {{{\text{t}}^2} + {{\left( {{{\text{r}}_1} - {{\text{r}}_2}} \right)}^2}} $ where d is the distance between the centers of two circles with radii ${{\text{r}}_1}$ and${{\text{r}}_2}$, and t is the length of the common tangent and we’ll get the same answer. This formula is used when the circles have a direct common tangent. On putting the given values we get,
$ \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)-
![seo images](https://www.vedantu.com/question-sets/7c35cd34-d479-44ae-9a2d-3379009d030d2421250251965047030.png)
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)