Two circles touch internally. The sum of their areas is \[116\pi cm{}^\text{2}\]and the distance between their centres is 6 cm. Find the radius of circles.
A) 8cm, 20cm
B) 4cm, 10cm
C) 6cm, 8cm
D) 5cm, 9cm
Last updated date: 18th Mar 2023
•
Total views: 303.6k
•
Views today: 5.82k
Answer
303.6k+ views
Hint: To solve the question, we have to represent the given data in a diagram which hints the relation between the radius of the circles. Apply relevant formulae to solve the equation and find out the values.
Complete step-by-step answer:
Let the radius of two circles be \[x,y\] where \[x>y\].
The distance between the centre of the circles given = 6 cm.
From the diagram we observe that the distance between the centre of the circles = x - y
\[\Rightarrow x-y=6\] ……. (1)
We know that the formula of area of the circle \[=\pi {{r}^{2}}\]
Where r is the radius of the circle.
Thus, the sum of the areas of the circles with radius x, y \[=\pi {{x}^{2}}+\pi {{y}^{2}}\].
The sum of the areas of the circles given \[=116\pi cm{}^\text{2}\].
\[\Rightarrow \pi {{x}^{2}}+\pi {{y}^{2}}=116\pi cm{}^\text{2}\]
\[\pi \left( {{x}^{2}}+{{y}^{2}} \right)=116\pi \]
\[{{x}^{2}}+{{y}^{2}}=116\]
By substituting the equation (1) in the above equation, we get
\[{{\left( 6+y \right)}^{2}}+{{y}^{2}}=116\] ……. (2)
We know that the formula for \[{{\left( a+b \right)}^{2}}\] is given by \[{{a}^{2}}+{{b}^{2}}+2ab\]
On comparing the expression \[{{\left( a+b \right)}^{2}}\]with \[{{\left( 6+y \right)}^{2}}\], we get a = 6 and b = y
By substituting the above formula in equation (2), we get that
\[{{6}^{2}}+2(6)y+{{y}^{2}}+{{y}^{2}}=116\]
\[36+12y+{{y}^{2}}+{{y}^{2}}=116\]
\[2{{y}^{2}}+12y-116+36=0\]
\[2{{y}^{2}}+12y-80=0\]
\[2\left( {{y}^{2}}+6y-40 \right)=0\]
\[{{y}^{2}}+6y-40=0\]
\[{{y}^{2}}+10y-4y-40=0\]
\[\left( y+10 \right)\left( y-4 \right)=0\]
Thus, the possible values of y = 4 cm, -10 cm.
Radius is a measurement which cannot be negative.
\[\Rightarrow \]y = 4cm
The value of x = 6 + y = 6 + 4 = 10cm.
Thus, the radius of the circles are 4 cm,10 cm.
Hence the option (b) is the correct answer.
Note: The possibility of mistake is the calculations while solving the quadratic equations. The alternative method to solve the question is by option elimination method, from the given information we understand that the difference between the radius of the circles is equal to 6 cm. Thus, we can arrive at the solution while eliminating the other three options.
Complete step-by-step answer:
Let the radius of two circles be \[x,y\] where \[x>y\].

The distance between the centre of the circles given = 6 cm.
From the diagram we observe that the distance between the centre of the circles = x - y
\[\Rightarrow x-y=6\] ……. (1)
We know that the formula of area of the circle \[=\pi {{r}^{2}}\]
Where r is the radius of the circle.
Thus, the sum of the areas of the circles with radius x, y \[=\pi {{x}^{2}}+\pi {{y}^{2}}\].
The sum of the areas of the circles given \[=116\pi cm{}^\text{2}\].
\[\Rightarrow \pi {{x}^{2}}+\pi {{y}^{2}}=116\pi cm{}^\text{2}\]
\[\pi \left( {{x}^{2}}+{{y}^{2}} \right)=116\pi \]
\[{{x}^{2}}+{{y}^{2}}=116\]
By substituting the equation (1) in the above equation, we get
\[{{\left( 6+y \right)}^{2}}+{{y}^{2}}=116\] ……. (2)
We know that the formula for \[{{\left( a+b \right)}^{2}}\] is given by \[{{a}^{2}}+{{b}^{2}}+2ab\]
On comparing the expression \[{{\left( a+b \right)}^{2}}\]with \[{{\left( 6+y \right)}^{2}}\], we get a = 6 and b = y
By substituting the above formula in equation (2), we get that
\[{{6}^{2}}+2(6)y+{{y}^{2}}+{{y}^{2}}=116\]
\[36+12y+{{y}^{2}}+{{y}^{2}}=116\]
\[2{{y}^{2}}+12y-116+36=0\]
\[2{{y}^{2}}+12y-80=0\]
\[2\left( {{y}^{2}}+6y-40 \right)=0\]
\[{{y}^{2}}+6y-40=0\]
\[{{y}^{2}}+10y-4y-40=0\]
\[\left( y+10 \right)\left( y-4 \right)=0\]
Thus, the possible values of y = 4 cm, -10 cm.
Radius is a measurement which cannot be negative.
\[\Rightarrow \]y = 4cm
The value of x = 6 + y = 6 + 4 = 10cm.
Thus, the radius of the circles are 4 cm,10 cm.
Hence the option (b) is the correct answer.
Note: The possibility of mistake is the calculations while solving the quadratic equations. The alternative method to solve the question is by option elimination method, from the given information we understand that the difference between the radius of the circles is equal to 6 cm. Thus, we can arrive at the solution while eliminating the other three options.
Recently Updated Pages
Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE
