Question

A circle is given with centre C and radius R > 0 Fixed points S and S, lie on a diameter line of this circle and are symmetric about C. Variable line segments SM are drawn so that M lies on the circle. SM is produced to H so that SM-MH. P is the point of intersection of line SH and the perpendicular bisector of segment SH. Match the conditions given in column I with the corresponding nature of locus of point P given in column II. (only single matching to be done). Column I A) S and S, do not coincide and lie inside the circle B) S and S, coincide with CS C) S and S, do not coincide and lie on the circle D) S and S, do not coincide and lie outside the circle

Column II (P) Empty set (Q) Single point (R) Straight line (S) Circle (T) Ellipse (U) Hyperbola (V) None of these

• A. A-S, B-S, C-V, D-S
• B. A-S, B-P, C-V, D-S
• C. A-S, B-S, C-Q, D-S
• D. A-S, B-S, C-V, D-S

Answer 1

Answer: (D)

A-S, B-S, C-V, D-S

Answer 2

Let C be the origin (0,0). The circle is given by $x^2 + y^2 = R^2$. Let the diameter line be the x-axis. Then S and S' are at $(s, 0)$ and $(-s, 0)$ respectively, where $s \ge 0$.

M is a point on the circle, so $M = (R\cos\theta, R\sin\theta)$. SM is produced to H such that SM = MH. This means M is the midpoint of SH. If $S = (s_x, s_y)$ and $H = (h_x, h_y)$, then $M = \left(\frac{s_x+h_x}{2}, \frac{s_y+h_y}{2}\right)$. So, $h_x = 2x_M - s_x$ and $h_y = 2y_M - s_y$. If $S = (s, 0)$ and $M = (x_M, y_M)$, then $H = (2x_M - s, 2y_M)$.

P is the point of intersection of line SH and the perpendicular bisector of segment SH. The line SH passes through S and H. The midpoint of SH is M. The perpendicular bisector of SH is the line passing through M and perpendicular to SH. Since P lies on the line SH and also on the perpendicular bisector of SH (which passes through M and is perpendicular to SH), P must be the point M itself. Therefore, the locus of P is the same as the locus of M, which is the circle.

We need to check the conditions under which P is well-defined. P is well-defined as long as the line SH and its perpendicular bisector are well-defined. This is true unless S coincides with M.

Case A: S and S' do not coincide and lie inside the circle. Here, $0 < |s| < R$. S is not on the circle. Thus, M can never coincide with S. P is always well-defined and P = M. The locus of P is the circle.

Case B: S and S' coincide with C. Here, $s = 0$. So S = (0, 0). M is on the circle. M cannot coincide with S unless R=0, which is not allowed. P is always well-defined and P = M. The locus of P is the circle.

Case C: S and S' do not coincide and lie on the circle. Here, $|s| = R$ and $s \neq 0$. Let S be $(R, 0)$. If M is chosen as $(R, 0)$ (i.e., M = S), then S = M. In this case, S = M = H, and the line SH is not uniquely defined, nor is its perpendicular bisector. Thus, P is undefined for this specific point M. For all other points M on the circle, P = M. The locus of P is the circle excluding the point S. Since "circle excluding a point" is not one of the options (P-V), this case corresponds to (V) None of these.

Case D: S and S' do not coincide and lie outside the circle. Here, $|s| > R$ and $s \neq 0$. S is outside the circle. Thus, M can never coincide with S. P is always well-defined and P = M. The locus of P is the circle.

Summary: A) Locus of P is a Circle (S). B) Locus of P is a Circle (S). C) Locus of P is not a standard shape from the options, hence None of these (V). D) Locus of P is a Circle (S).

The matching is A-S, B-S, C-V, D-S.