Question

In the given figure, circle with centre M touches the circle with centre N at point T. Radius RM touches the smaller circle at S. Radii of circles are 9 cm and 2.5 cm. Find the answers to the following questions hence find the ratio MS:SR.
(1) Find the length of segment MT
(2) Find the length of seg MN
(3) Find the measure of ∠NSM. 

Answer 1

Radius of circle with centre M = 9 cm
Radius of circle with centre N = 2.5 cm
Join MT and NS. 

If two circles touch each other, their point of contact lie on the line joining their centres. So, the points M, N and T are collinear.
(1)
Length of segment MT = 9 cm           (Radius of circle with centre M)
Thus, the length of the segment MT is 9 cm.
(2)
Length of segment NT = 2.5 cm         (Radius of circle with centre N)
∴ Length of segment MN = Length of segment MT − Length of segment NT = 9 − 2.5 = 6.5 cm
Thus, the length of the segment MN is 6.5 cm.
(3)
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
In the given figure, seg RM is tangent to the circle with centre N at point S.
∴ ∠NSM = 90º
In right ∆NSM,

\[{MN}^2 = {NS}^2 + {SM}^2 \]
\[ \Rightarrow SM = \sqrt{{MN}^2 - {NS}^2}\]
\[ \Rightarrow SM = \sqrt{\left( 6 . 5 \right)^2 - \left( 2 . 5 \right)^2}\]
\[ \Rightarrow SM = \sqrt{42 . 25 - 6 . 25}\]
\[ \Rightarrow SM = \sqrt{36} = 6 \] cm

∴ SR = MR − SM = 9 − 6 = 3 cm            (MR = Radius of the circle with centre M)
⇒ MS : SR = 6 cm : 3 cm = 2 : 1
Thus, the ratio MS : SR is 2 : 1.