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Find the locus of the midpoint of the chord of the circle $x^2 + y^2=a^2$ which subtends a $90°$ angle at point $(p,q)$ lying inside the circle.

I tried to solve it by taking that let the chord intersect the circle at $(x_1,y_1)$ and $(x_2,y_2)$. Then I found out their slopes and took their product as $-1$. I also tried by taking the lines joining center from chord as perpendicular.

But I couldn't do it. Please tell me a way.

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  • $\begingroup$ Can someone please suggest a solution? $\endgroup$
    – Ava
    Commented Jan 14, 2016 at 4:08
  • $\begingroup$ It appears that the locus is a circle -- but I am not sure why (yet). I will give it some thought. $\endgroup$
    – mweiss
    Commented Jan 14, 2016 at 4:09

3 Answers 3

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WLOG, let the two extreme points of the chord be $A(a\cos u,a\sin u), B(a\cos v,a\sin v)$

If $P(h,k)$ the midpoint of $A,B$

$2h=a(\cos u+\cos v)\ \ \ \ (1)$

$2k=a(\sin u+\sin v)\ \ \ \ (2)$

$\implies4(h^2+k^2)=a^2\{2+2\cos(u-v)\}\iff\cos(u-v)=?\ \ \ \ (3)$

Now, $AB^2=PA^2+PB^2$

$$\implies4a^2\{(\cos u-\cos v)^2+(\sin u-\sin v)^2\}$$ $$=(p-2a\cos u)^2+(q-2a\sin u)^2+(p-2a\cos v)^2+(q-2a\sin v)^2$$

$\iff4a^2\{2-2\underbrace{\cos(u-v)}\}=2(p^2+q^2)$ $-2p\cdot2\underbrace{a(\cos u+\cos v)}-2q\cdot2\underbrace{a(\sin u+\sin v)}$

Use $(1),(2),(3)$ for the under-braced parts.

Can you take it from here?

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  • $\begingroup$ What are $ u,v ? $\endgroup$
    – Narasimham
    Commented Jan 13, 2016 at 16:57
  • $\begingroup$ @Narasimham,See mathworld.wolfram.com/ParametricEquations.html $\endgroup$ Commented Jan 13, 2016 at 16:58
  • $\begingroup$ You are taking u+v=90 ? $\endgroup$
    – Ava
    Commented Jan 13, 2016 at 17:25
  • $\begingroup$ @Ava, why do u think so? $\endgroup$ Commented Jan 13, 2016 at 17:29
  • $\begingroup$ u haven't mentioned exactly that the angle u and v are made by which lines.. are they by the end points of the chord or something else and also, the angle is made with the origin or point (p,q)? $\endgroup$
    – Ava
    Commented Jan 13, 2016 at 17:32
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There is another way to solve the problem; Given the circle equation is... $x^2 + y^2 -a^2$=0;

Let the mid point of the chord be $(h,k)$ Equation of the chord when mid point is given is $ T=S_1$. $hx+ky-a^2=h^2+k^2-a^2$ or $$hx+ky-h^2-k^2=0$$

The equation of all cirles passing through the intersection point of the circle and the chord will be

$$x^2+y^2-a^2+\lambda(hx+ky-h^2-k^2)=0$$ Now we have to select a circle from the above equation that should satisfy the given condition. The circle should subtend right angle by the chord at the point $(p,q)$

This will be possible when the circle passes through the point and when the chord is diameter of the circle. Center of the unique circle should be $(h,k)$ and should pass through the point $(p,q)$.

$$x^2+y^2-a^2+\lambda(hx+ky-h^2-k^2)=0$$ expand $x^2 + y^2 +\lambda hx+ \lambda ky....$ This imply the center of the circle is $(-\lambda h/2,-\lambda k/2)$ which should be equal to $(h,k)$. Thus $\lambda$ is -2. So, $$x^2+y^2-a^2+(-2)(hx+ky-h^2-k^2)=0$$ and $x=p$ and $y=q$ since it passes through $(p,q)$ Substitute and simplify $$p^2 +q^2 -a^2-2hp+-2kq+2h^2+2k^2=0$$

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In the linked dynamic diagram we see a circle (centered at $O$) and an interior point $P$. Two perpendicular lines (shown as dashed in the diagram) intersect at $P$, and intercept the circle in four points, which are joined by four chords (shown as boldfaced segments). So each of the four boldfaced chords subtends a $90°$ angle at $P$. The midpoints of the four chords are $W,X,Y$ and $Z$, shown in red.

The orientation of the two perpendicular lines can be changed by dragging point $Q$ (purple) around the circumference of the circle. As $Q$ moves, the chords and their midpoints move as well. So the question is:

As $Q$ varies, how do the red points move? What path is traced out by $WX,Y$ and $Z$?

The “Show/Hide Locus” toggle button reveals the path. It appears to be a circle! More precisely, it appears to be a circle centered at the midpoint $M$ of the segment joining $O$ to $P$. You can reveal point $M$ (shown in green) with the second toggle button.

I don’t, unfortunately, have a proof for you of why this is so, but now at least you know what you need to prove.

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