Question

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. a circle of radius $\frac{r}{2}$
• B. a circle of radius 2r
• C. a parabola of latus rectum 4r
• D. a rectangular hyperbola

Answer 1

The correct option is A a circle of radius $\frac{r}{2}$

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

$A{C}^2=A{R}^2+C{R}^2⟹h^2+k^2=2P^2{r}^2/16+r^2/16=2r/8P⟹r^2/8=r/2P\therefore P=r/2$

$\therefore$ The locus of $P=r/2$ i.e. a circle of radius $r/2$