Question

Let P and Q be two distinct points on a circle which has center at C(2, 3) and which passes through origin O. If OC is perpendicular to both the line segments CP and CQ, then the set\{P, Q\} is equal to :

• A. $\{(2+2\sqrt{2}, 3+\sqrt{5}), (2-2\sqrt{2}, 3-\sqrt{5})\}$
• B. $\{(2+2\sqrt{2}, 3-\sqrt{5}), (2-2\sqrt{2}, 3+\sqrt{5})\}$
• C. $\{(-1, 5), (5, 1)\}$
• D. $\{(4, 0), (0, 6)\}$

Hint:

When two vectors are perpendicular, their dot product is zero. This is a powerful tool to create a linear equation relating the coordinates of an unknown point. Combine this with the equation of the locus (like a circle) to solve for the point(s).

Answer 1

The Correct Option is C

First, find the equation of the circle. The center is C(2, 3). It passes through the origin O(0, 0).
The radius 'r' is the distance OC.
$r = \sqrt{(2-0)^2 + (3-0)^2} = \sqrt{4+9} = \sqrt{13}$.
The equation of the circle is $(x-2)^2 + (y-3)^2 = (\sqrt{13})^2 = 13$.
$x^2 - 4x + 4 + y^2 - 6y + 9 = 13 \implies x^2 + y^2 - 4x - 6y = 0$.
The line passing through P and Q is perpendicular to OC and passes through the center C.
The vector $\vec{OC}$ is $(2-0)\hat{i} + (3-0)\hat{j} = 2\hat{i} + 3\hat{j}$.
The slope of OC is $m_{OC} = 3/2$.
The line PQ is perpendicular to OC. The slope of PQ is $m_{PQ} = -1/m_{OC} = -2/3$.
The problem states OC is perpendicular to CP. This means the line passing through P and Q is perpendicular to OC, but passes through C. Wait, no. OC is perp to CP. This means the vector dot product is zero. Let P=(x,y).
Vector $\vec{CP} = (x-2)\hat{i} + (y-3)\hat{j}$.
Given $\vec{OC} \perp \vec{CP}$, so their dot product is zero.
$\vec{OC} \cdot \vec{CP} = (2)(x-2) + (3)(y-3) = 0$.
$2x - 4 + 3y - 9 = 0 \implies 2x + 3y = 13$.
This is the equation of the line on which both points P and Q lie.
To find P and Q, we solve the system of equations for the circle and the line.
From $2x + 3y = 13$, we get $x = \frac{13-3y}{2}$.
Substitute this into the circle equation $x^2 + y^2 - 4x - 6y = 0$:
$\left(\frac{13-3y}{2}\right)^2 + y^2 - 4\left(\frac{13-3y}{2}\right) - 6y = 0$.
$\frac{169 - 78y + 9y^2}{4} + y^2 - 2(13-3y) - 6y = 0$.
$169 - 78y + 9y^2 + 4y^2 - 104 + 24y - 24y = 0$.
$13y^2 - 78y + 65 = 0$.
Divide by 13: $y^2 - 6y + 5 = 0$.
Factor the quadratic: $(y-1)(y-5) = 0$.
So, $y=1$ or $y=5$.
If $y=1$, then $x = \frac{13-3(1)}{2} = \frac{10}{2} = 5$. The point is (5, 1).
If $y=5$, then $x = \frac{13-3(5)}{2} = \frac{-2}{2} = -1$. The point is (-1, 5).
The set of points \{P, Q\} is $\{(-1, 5), (5, 1)\}$.