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P is the variable point on the circle with centre at C. CA and CB are perpendiculars from C on x-axis and y-axis, respectively, then the locus of the centroid of triangle PAB is.

Option: 1

Parabola


Option: 2

Ellipse


Option: 3

Straight Line


Option: 4

Circle


Answers (1)

best_answer

Let the circle be \mathrm{x^2+y^2+2 g x+2 f y+c=0}

Let the centroid of triangle PAB be (h, k).
\mathrm{ \therefore \quad 3 h=-2 g+r \cos \theta }
\mathrm{ \text { and } 3 k=-2 f+r \sin \theta }

\mathrm{ \Rightarrow(3 h+2 g)^2+(3 k+2 f)^2=r^2 }
\mathrm{ \Rightarrow(x+2 g / 3)^2+(y+2 f / 3)^2=r^2 / 9 }
This is the required locus which is a circle.

Posted by

jitender.kumar

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