Question

Let $S: x^2 + y^2 - 8x - 6y + 24 = 0$ be a circle and O is the origin. Let OAB is the line intersecting the circle at A and B. On the chord AB a point P is taken. The locus of the point P in each of the following cases.

(i) OP is the arithmetic mean of OA and OB

(ii) OP is the geometric mean of OA and OB

(iii) OP is the harmonic mean between OA and OB

• A. (i) $(x-2)^2 + (y-3/2)^2 = 25/4$, (ii) $x^2 + y^2 = 24$, (iii) $4x + 3y = 24$
• B. (i) $x^2 + y^2 = 24$, (ii) $(x-2)^2 + (y-3/2)^2 = 25/4$, (iii) $4x + 3y = 24$
• C. (i) $4x + 3y = 24$, (ii) $x^2 + y^2 = 24$, (iii) $(x-2)^2 + (y-3/2)^2 = 25/4$
• D. (i) $(x-2)^2 + (y-3/2)^2 = 25/4$, (ii) $4x + 3y = 24$, (iii) $x^2 + y^2 = 24$

Answer 1

Answer: (A)

(i) $(x-2)^2 + (y-3/2)^2 = 25/4$ (ii) $x^2 + y^2 = 24$ (iii) $4x + 3y = 24$

Answer 2

The equation of the circle $S$ is $x^2 + y^2 - 8x - 6y + 24 = 0$. Completing the square gives $(x-4)^2 + (y-3)^2 = 1$, which is a circle with center $C(4, 3)$ and radius $r=1$. The origin $O(0,0)$ is outside the circle since $OC=5 > r$.

A line OAB passes through the origin. Let the line be represented in polar coordinates as $d(\cos\theta, \sin\theta)$, where $d$ is the distance from the origin. Substituting into the circle equation: $d^2 - d(8\cos\theta + 6\sin\theta) + 24 = 0$. The roots of this quadratic in $d$ are $OA$ and $OB$. By Vieta's formulas: $OA \cdot OB = 24$ $OA + OB = 8\cos\theta + 6\sin\theta$

Let $OP = d_P$. P is on the line OAB, so its coordinates are $(d_P\cos\theta, d_P\sin\theta)$.

(i) OP is the arithmetic mean of OA and OB: $d_P = \frac{OA + OB}{2} = \frac{8\cos\theta + 6\sin\theta}{2} = 4\cos\theta + 3\sin\theta$. Substituting $x = d_P\cos\theta$ and $y = d_P\sin\theta$: $\sqrt{x^2+y^2} = 4\left(\frac{x}{\sqrt{x^2+y^2}}\right) + 3\left(\frac{y}{\sqrt{x^2+y^2}}\right)$ $x^2+y^2 = 4x + 3y$ $(x-2)^2 + (y-3/2)^2 = 4 + 9/4 = 25/4$. This is a circle with center $(2, 3/2)$ and radius $5/2$.

(ii) OP is the geometric mean of OA and OB: $d_P = \sqrt{OA \cdot OB} = \sqrt{24}$. $d_P^2 = 24$. $x^2+y^2 = 24$. This is a circle centered at the origin with radius $\sqrt{24} = 2\sqrt{6}$.

(iii) OP is the harmonic mean between OA and OB: $d_P = \frac{2}{\frac{1}{OA} + \frac{1}{OB}} = \frac{2 OA \cdot OB}{OA + OB} = \frac{2 \cdot 24}{8\cos\theta + 6\sin\theta} = \frac{48}{8\cos\theta + 6\sin\theta}$. $d_P(8\cos\theta + 6\sin\theta) = 48$. Substituting $x = d_P\cos\theta$ and $y = d_P\sin\theta$: $8x + 6y = 48$, which simplifies to $4x + 3y = 24$. This is a straight line.