23. Point M lies in the exterior of
a circle with centre a and a tangent from M touches the circle at n
if AM = 41 cm and MN = 40 cm, find the radius of the circle.
Answers
If the tangent MN = 40 cm and the distance from centre A to point M is 41 cm then the radius of the circle is 9 cm.
Step-by-step explanation:
It is given,
A tangent from point M exterior to the circle touches the circle at N and MN = 40 cm and the distance from the center A to point M is 41 cm.
We know that a tangent to a circle is perpendicular to the radius through the point of contact, therefore,
∠ANM = 90°
Let the radius of the circle be denoted by “AN”.
Now referring to the figure attached below, we will use Pythagoras theorem, for ∆ AMN,
AM² = MN² + AN²
Rearranging the theorem
⇒ AN² = AM² - MN²
⇒ AN² = 41² - 40² ….. [substituting the given values]
⇒ AN = √[41² - 40²]
⇒ AN = √[1681 - 1600]
⇒ AN = √[81]
⇒ AN = 9 cm
Thus, the radius of the circle is 9 cm.
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Step-by-step explanation:
here we have to find r-?
We have given that mn=40 and am=41
In ∆MAN
MN^2=AN^2+AM^2
40^2=AN^2+41^2
1600=AN^2+1681
AN^2=1681-1600
AN^2=81
AN=9