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Let the base AB of a triangle ABC be fixed and the vertex C lies on a fixed circle of radius r. Lines through A and B are drawn to intersect CB and CA, respectively, at E and F such that CE: EB=1:2 and CF: FA= 1:2.If the point of intersection P of these lines lies on the median through AB for all positions of AB then the locus of P is.
a circle of radius $\frac{r}{2}$
a circle of radius 2r
a parabola of latus rectum 4r
a rectangular hyperbola

A

a circle of radius

\frac{r}{2}

B

a circle of radius 2r

C

a rectangular hyperbola

D

a parabola of latus rectum 4r

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Solution

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Consider the circle in which the triangle is inscribed with bare AB. The vertex C of $△$ ABC is fixed at the centre of the circle of radius r.
A line is drawn through A and it intersects CB at E and dividing CB in the ratio CE: EB = 1:2. Similarly, a line is drawn through B and it intersects AC at F and divides AC in the ratio 1:2. Let P be the centroid i.e. P (h, k)
Consider $△$ ACR
$A C^{2} = A R^{2} + C R^{2} ⟹ h^{2} + k^{2} = 2 P^{2} r^{2} / 16 + r^{2} / 16 = 2 r / 8 P ⟹ r^{2} / 8 = r / 2 P ∴ P = r / 2$
$∴$ The locus of $P = r / 2$ i.e. a circle of radius $r / 2$

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