Question

From $(3,4)$ chords are drawn to the circle $x^2+y^2-4x=0$. The locus of the mid points of the chords is:

• A. $x^2+y^2-5x-4y+6=0$
• B. $x^2+y^2+5x-4y+6=0$
• C. $x^2+y^2-5x+4y+6=0$
• D. $x^2+y^2-5x-4y-6=0$

Answer 1

The correct option is A $x^2+y^2-5x-4y+6=0$

Consider the given equation.

$x^2+y^2-4x=0$,

Add $4$ to both sides.

$x^2-4x+4+y^2=4$

Thus, the given circle is $(x-2{)}^2+y²=4$ so has centre $C$ at $(2,0)$ and radius $2$.

Say $A(3,4)$ and let any chord through $A$ have mid-point $B$.
Then $CB$ is perpendicular to chord and $\angle ABC=90°$
∴ $B$ lies on circle whose diameter is $AC$. (angle in semicircle$=90°$)
Centre of locus circle $=\frac{1}{2}\left[\right(3,4)+(2,0\left)\right]=(\frac{5}{2},2)$

Now,

Radius $R^2={(\frac{5}{2}–2)}^2+(2–0{)}^2=\frac{17}{4}$

Thus,

The locus of the mid points of the chords is:

${(x-\frac{5}{2})}^2+(y-2{)}^2=\frac{17}{4}$

$x^2+y^2-5x-4y+6=0$.

Note : For real intersections the locus is restricted to that part of the this circle that lies within the given circle. However, imaginary intersections still have a real mid-point and these account for the rest of the locus circle.