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.

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