SATHEE: Circle 5 Question 18

Circle 5 Question 18

18. $C_{1}$ and $C_{2}$ are two concentric circles, the radius of $C_{2}$ being twice that of $C_{1}$ . From a point $P$ on $C_{2}$ , tangents $P A$ and $P B$ are drawn to $C_{1}$ . Prove that the centroid of the $△ P A B$ lies on $C_{1}$ .

$(1998, 8$ M)

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Solution:

Let the coordinate of point $P$ be $(2 r \cos θ, 2 r \sin θ)$

We have, $O A = r, O P = 2 r$

Since, $△ O A P$ is a right angled triangle.

Circle 415

$\cos φ = 1 / 2 \Rightarrow φ = π / 3$

$∴$ Coordinates of $A$ are $r \cos (θ - π / 3), r \sin (θ - π / 3)$

and that of $B$ are $[r \cos (θ + π / 3), r \sin (θ + π / 3)]$

If $p, q$ is the centroid of $△ P A B$ , then $p = \frac{1}{3} [r \cos (θ - π / 3) + r \cos (θ + π / 3) + 2 r \cos θ]$

$= \frac{1}{3} [r \cos (θ - π / 3) + \cos (θ + π / 3) + 2 r \cos θ]$

$= \frac{1}{3} r 2 \cos \frac{θ - \frac{π}{3} + θ + \frac{π}{3}}{2} \cdot \cos \frac{θ - \frac{π}{3} - θ - \frac{π}{3}}{2} + 2 r \cos θ$

$= \frac{1}{3} [r 2 \cos θ \cos π / 3 + 2 r \cos θ]$

$= \frac{1}{3} [r \cdot \cos θ + 2 r \cos θ] = r \cos θ$

and $q = \frac{1}{3} [r \sin θ - \frac{π}{3} + r \sin θ + \frac{π}{3} + 2 r \sin θ]$

$= \frac{1}{3} [r \sin θ - \frac{π}{3} + \sin θ + \frac{π}{3} + 2 r \sin θ]$

$= \frac{1}{3} r 2 \sin \frac{θ - \frac{π}{3} + θ + \frac{π}{3}}{2} \cdot \cos \frac{θ - \frac{π}{3} - θ - \frac{π}{3}}{2} + 2 r \sin θ$

$= \frac{1}{3} [r (2 \sin θ \cos π / 3) + 2 r \sin θ]$

$= \frac{1}{3} [r (\sin θ) + 2 r \sin θ] = r \sin θ$

Now, $(p, q) = (r \cos θ, r \sin θ)$ lies on $x^{2} + y^{2} = r^{2}$ which is called $C_{1}$ .

Circle 5 Question 18

18. $C_{1}$ and $C_{2}$ are two concentric circles, the radius of $C_{2}$ being twice that of $C_{1}$ . From a point $P$ on $C_{2}$ , tangents $P A$ and $P B$ are drawn to $C_{1}$ . Prove that the centroid of the $△ P A B$ lies on $C_{1}$ .

Solution:

Circle 415

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Circle 5 Question 18

18. C1 and C2 are two concentric circles, the radius of C2 being twice that of C1. From a point P on C2, tangents PA and PB are drawn to C1. Prove that the centroid of the △PAB lies on C1.

Solution:

Circle 415

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

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NCERT Exercise Video Solution

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18. $C_{1}$ and $C_{2}$ are two concentric circles, the radius of $C_{2}$ being twice that of $C_{1}$ . From a point $P$ on $C_{2}$ , tangents $P A$ and $P B$ are drawn to $C_{1}$ . Prove that the centroid of the $△ P A B$ lies on $C_{1}$ .