Question

a circle of constant radius 3k passes through (0,0) and cuts the axes in A and B then locus of centriod of triangle OAB is

Answer 1

Answer:

Refer to the rough diagram for reference.

It is given that a circle cuts the x & y axes at two different points, namely 'A' & 'B'.

Let us assume, it cuts the 'x' axis at A. Hence the coordinates of 'A' would be:

• A = ( x, 0 )

Similarly, it cuts the 'y' axis at B. Hence the coordinates of 'B' would be:

• B = ( 0, y )

Therefore on joining the points O, A and B we get a right angled triangle.

Now on close observation, we see that, AB is the diameter of the circle.

Therefore the length of AB = 2 × Radius of the circle.

It is given that the radius is 3k units.

Hence AB = 2 × 3k = 6k.

Applying Pythagoras Theorem in Δ OAB, we get:

⇒ AO² + OB² = AB²

⇒ x² + y² = ( 6k )²

⇒ x² + y² = 36 k²   ...(1)

Now, we know that the formula to calculate centroid of a triangle is:

$\boxed{ \bf{ \textbf{Centroid (a,b) } = \dfrac{x_1 + x_2+x_3}{3}\:,\:\dfrac{y_1+y_2+y_3}{3}}}$

According to our question,

• x₁ = 0, x₂ = x, x₃ = 0
• y₁ = 0, y₂ = 0, y₃ = y

Substituting the values we get:

$\implies (a,b) = \dfrac{0+x+0}{3}\:,\: \dfrac{0+0+y}{3}\\\\\\\implies \boxed{ \bf{ (a,b) = \dfrac{x}{3}\:,\:\dfrac{y}{3} }}$

Therefore we get:

⇒ x = 3a & y = 3b    ...(2)

Substituting (2) in (1) we get:

⇒ (3a)² + (3b)² = 36 k²

⇒ 9a² + 9b² = 36 k²

⇒ 9 ( a² + b² ) = 36 k²

⇒ a² + b² = 4 k²

Rewriting "a as x" and "b as y" we get:

⇒ x² + y² = 4k²

Hence the locus of centroid (a,b) is x² + y² = 4k²