Question

From (3, 4) chords are drawn to the circle x² + y² - 4x = 0. The locus of the mid points of the chords is :

• A. x² + y² - 5x – 4y + 6 = 0
• B. x² + y² + 5x – 4y + 6 = 0
• C. x² + y² - 5x + 4y + 6 = 0
• D. x² + y²- 5x – 4y - 6 = 0

Answer 1

Answer: (A)

x² + y² - 5x – 4y + 6 = 0

Answer 2

Let $P=(3,4)$ be the given point. The equation of the circle is $x^2 + y^2 - 4x = 0$. Completing the square, we get $(x-2)^2 + y^2 = 2^2$. The center of the circle is $O=(2,0)$ and the radius is $r=2$.

Let $M(h, k)$ be the midpoint of a chord drawn from point $P(3, 4)$ to a point on the circle. The line segment $OM$ is perpendicular to the chord. Since $PM$ is part of the chord, $OM \perp PM$.

The slope of $OM$ is $m_{OM} = \frac{k-0}{h-2} = \frac{k}{h-2}$. The slope of $PM$ is $m_{PM} = \frac{k-4}{h-3}$.

Since $OM \perp PM$, the product of their slopes is $-1$: $m_{OM} \cdot m_{PM} = -1$ $\left(\frac{k}{h-2}\right) \cdot \left(\frac{k-4}{h-3}\right) = -1$ $k(k-4) = -(h-2)(h-3)$ $k^2 - 4k = -(h^2 - 5h + 6)$ $k^2 - 4k = -h^2 + 5h - 6$

Rearranging the terms, we get the locus equation: $h^2 + k^2 - 5h - 4k + 6 = 0$

Replacing $(h, k)$ with $(x, y)$ to represent the locus of the midpoints: $x^2 + y^2 - 5x - 4y + 6 = 0$