Two circles touch externally.the sum of their areas is 117pi cm2 and the distance between their centres is 15cm. Find the radii of the two circle
Answers
Let radius of circle AA be r_{1}r
1
and radius of circle BB be r_{2}r
2
Given, r_{1}+r_{2}=15r
1
+r
2
=15 -------(i)
and \pi{r_{1}}^2+\pi{r_{2}}^2=117\piπr
1
2
+πr
2
2
=117π
\Rightarrow {r_{1}}^2+{r_{2}}^2=117⇒r
1
2
+r
2
2
=117 ------(ii)
(i)\Rightarrow r_{1}+r_{2}=15⇒r
1
+r
2
=15
\Rightarrow r_{1}=15-r_{2}⇒ r
1
=15−r
2
Now, putting this value in (ii), we get
\Rightarrow (15-r_{2})^2+{r_{2}}^2=117⇒(15−r
2
)
2
+r
2
2
=117
\Rightarrow 15^2-2\times 15 \times r_{2}+{r_{2}}^2+{r_{2}}^2=117⇒15
2
−2×15×r
2
+r
2
2
+r
2
2
=117
\Rightarrow 225-30r_{2}+2{r_{2}}^2=117⇒225−30r
2
+2r
2
2
=117
\Rightarrow 2{r_{2}}^2-30r_{2}+108=0⇒2r
2
2
−30r
2
+108=0
\Rightarrow2{r_{2}}^2-18r_{2}-12r_{2}+108=0⇒2r
2
2
−18r
2
−12r
2
+108=0
\Rightarrow 2r_{2}(r_{2}-9)-12(r_{2}-9)=0⇒2r
2
(r
2
−9)−12(r
2
−9)=0
\Rightarrow (r_{2}-9)(2r_{2}-12)=0⇒(r
2
−9)(2r
2
−12)=0
\Rightarrow r_{2}=9⇒r
2
=9 or r_{2}=\dfrac{12}{2}=6r
2
=
2
12
=6
\Rightarrow r_{1}=15-9⇒r
1
=15−9 or r_{1}=15-6r
1
=15−6
\Rightarrow r_{1}=6⇒r
1
=6 or r_{1}=9r
1
=9
\Rightarrow r_{1}>r_{2}⇒r
1
>r
2
Thus, the radius of the two circles are 99 cm and 66 cm.
Answer:
Let radius of circle
A
be
r
1
and radius of circle
B
be
r
2
Given,
r
1
+
r
2
=
15
-------(i)
and
π
r
1
2
+
π
r
2
2
=
117
π
⇒
r
1
2
+
r
2
2
=
117
------(ii)
(i)
⇒
r
1
+
r
2
=
15
⇒
r
1
=
15
−
r
2
Now, putting this value in (ii), we get
⇒
(
15
−
r
2
)
2
+
r
2
2
=
117
⇒
15
2
−
2
×
15
×
r
2
+
r
2
2
+
r
2
2
=
117
⇒
225
−
30
r
2
+
2
r
2
2
=
117
⇒
2
r
2
2
−
30
r
2
+
108
=
0
⇒
2
r
2
2
−
18
r
2
−
12
r
2
+
108
=
0
⇒
2
r
2
(
r
2
−
9
)
−
12
(
r
2
−
9
)
=
0
⇒
(
r
2
−
9
)
(
2
r
2
−
12
)
=
0
⇒
r
2
=
9
or
r
2
=
12
2
=
6
⇒
r
1
=
15
−
9
or
r
1
=
15
−
6
⇒
r
1
=
6
or
r
1
=
9
⇒
r
1
>
r
2
Thus, the radius of the two circles are
9
cm and
6
cm.
Step-by-step explanation: