This is an I.S.I. Entrance Solution

**Problem:**

P is a variable point on a circle C and Q is a fixed point on the outside of C. R is a point in PQ dividing it in the ratio p:q, where p> 0 and q > 0 are fixed. Then the locus of R is

(A) a circle; (B) an ellipse; (C) a circle if p = q and an ellipse otherwise; (C) none of the above curves;

**Discussion:**

Also see I.S.I. & C.M.I. Entrance Program

WLOG C be a unit circle centered at (0,0). Then ( P= (\cos(t), \sin (t)) ). Suppose Q = (a, b).

If R divides PQ in m: n ratio, then ( R = \left( \frac{ma + ncos(t)}{m+n} , \frac{mb+ nsin(t)}{m+n} \right) )

Then the distance of R from ( \left ( \frac{ma}{m+n}, \frac{mb}{m+n} \right) ) is ( \sqrt { \left( \frac{ma + ncos(t)}{m+n} – \frac{ma}{m+n} \right)^2 + \left( \frac{mb +nsin(t)}{m+n} – \frac{mb}{m+n} \right)^2} )

But this equals: ( \sqrt {\frac{m^2+ n^2} {(m+n)^2}} ) which is a constant. Hence R is at a constant distance from ( \left ( \frac{ma}{m+n}, \frac{mb}{m+n} \right) ) is a constant.

Hence R traces out a circle.

## Theoretical remark:

The parametric equation of a circle with unit radius centered at origin is ( \cos(t), \sin(t) ) . This is the key idea in this problem.

Another idea is: if PQ is divided in m : n ratio R, what is the coordinate of R. You will need ‘section formula’ to compute that (this formula is a consequence of similarity of triangles).

Wonderful solution, you people really are the best..another approach can be using complex number.

Good to hear from you. You may post your solution at https://www.cheenta.com/support

It is a latex enabled forum. We will be glad to read your solution