Question

Two circles touch externally. The sum of their areas is 130$\pi$ sq.cm. and the distance between their centers is 14cm. Find the radii of the circles.
A) \[11cm\] and \[3cm\]
B) \[1cm\] and \[3cm\]
C) \[11cm\] and \[13cm\]
D) None of these

Answer 1

Hint:
In the above question we are provided the conditions that two circles are touching externally and the sum of their areas is \[130\pi sq.cm.\] and the distance between their centers is 14cm.we are asked to the radii of the circles. Here using the condition the distance of the centers that is given to be \[14cm\] that this the sum of the radii of both the circles is \[14cm\] driving relation from the given condition and putting it in another condition that is the sum of the areas are given we can find the radius of the circles.

Complete step by step solution:
Let us consider two circles with \[{r_1}\] and \[{r_2}\] as their radii and the larger be having the radii as \[{r_1}\]

Now we are given that the distance between the centers of the circles is \[14cm\] that is-
\[{r_1} + {r_2} = 14cm\]
So we get the needed relation between the \[{r_1}\] and \[{r_2}\]
Now also the sum of the areas of the two circles is \[130\pi sq.cm.\] let us suppose the areas of the two circles be \[{A_1}\text{ and }{A_2}\] so the another relation that is between the areas of the two circles is as-
\[{A_1} + {A_2} = 130.\pi q.cm\]
Now the area of any circle with radius r is as-\[\pi {r^2}\]
Now the relation of the areas become
\[\pi {r^2}_1 + \pi {r^2}_2 = 130\pi sq.cm\]
Now taking like terms common and simplifying the terms we get –
\[\pi ({r^2}_1 + {r^2}_2) = 130\pi sq.cm\]----------------\[1\]
Now using the relation between the \[{r_1}\] and \[{r_2}\] we will find the value of \[{r_2}\] in the form of the \[{r_1}\] that is -
\[{r_2} = 14 - {r_1}\]

Now putting the value of the \[{r_2}\] from the above equation in the equation \[1\] that is
\[\pi {r^2}_1 + {(14 - {r^2}_1)^2} = 130\pi sq.cm\]
Now removing \[\pi \] from both sides the above equation becomes –
\[
\Rightarrow {r_1}^2 + 196 + {r_1}^2 - 28{r_1} = 130 \\
\Rightarrow 2{r_1}^2 - 28{r_1} + 66 = 0 \\
\Rightarrow {r_1}^2 - 14{r_1} + 33 = 0 \\
\Rightarrow {r_1}^2 - 11{r_1} - 3{r_1} + 33 = 0 \\
\Rightarrow ({r_1} - 11)({r_1} - 3) = 0 \\
 \]
Now from the above equation we get the
\[({r_1} - 11) = 0\]also \[({r_1} - 3) = 0\]
Now \[{r_1} = 11,3\] now putting \[{r_1}\] in the equation \[{r_1} + {r_2} = 14cm\] we get \[{r_2} = 11, 3\]
As depicted in the figure \[{r_1}\] and assumed earlier being the larger of the two \[{r_1} = 11\]\[cm\] and \[{r_2} = 3\]\[cm\] so from the above given options the option which matches the resultant is the option A that is A \[11cm\] and \[3cm\] So the correct option is A.

Note:
While solving these kind of the questions of two circles touching internally or externally search for the primary condition from that part and drawing a figure firstly will help you to understand what kind of condition is being formed out which would help to get the answer correctly generally when two circles touch internally the difference of the two radii of the circles gives the distance between them which becomes the primary condition in solving the question.