# Find the locus of the midpoint

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.

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

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?

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