Midpoint of chord of contact

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Question: The locus of the mid-point of the chord of contact of tangents drawn from points lying on the straight line $4x - 5y = 20$ to the circle $x^2 + y^2 = 9$ is:

a) $20(x^2 - y^2)- 36x + 45y = 0$

b) $20(x^2 + y^2)+ 36x - 45y = 0$

c) $36(x^2 + y^2)- 20x + 45y = 0$

d) $36(x^2 + y^2)+ 20x - 45y = 0$

How would I even start this question? Please give me a hint!

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Centre of circle is origin O. Get equation of line L' through O perpendicular to given line L, and hence intersection points of L' with circle and L. Hence check L does not intersect circle. The midpoint of the chord is inside the circle, so the locus is bounded, so it cannot be (a).

Locus passes through O (points at infinity on L give the chord as a diameter), and comes close to the intersection point P of L and L'. (b), (c), (d) are all circles, so locus is a circle. By symmetry its centre should lie on OP and be close to the midpoint of OP. It is easy to read off the coordinates of the centre of the circles (b), (c), (d) and hence to conclude which must be correct.

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On

None of a), b), c) or d) (as written) is the answer.

By using the dotted $3-4-5$ triangle in the picture below, you can capture a convenient point on the locus, then test it in the various equations. The point $E$ in this picture is at $(9/5,0)$, which happens to be on $20(x^2 - y^2)- 36x + 45y = 0$.

If you compute the point $F$ using a similar strategy, you'll find that this point is $(0,-9/4)$ is not on $a)$. However, both points lie on $20(x^2 + y^2)- 36x + 45y = 0$. I guess you typoed $a)$.

In the graph, the smaller circle is my suggested fix for a).

diagram of the situation

It's not very tough to work out what transformations like this look like. Take a look at this diagram, viewed as being in $\Bbb C$, where $A$ is the origin:

enter image description here

Since $\triangle ACZ$ and $\triangle ACE$ are similar, $|AE|=|AC|\frac{|AC|}{|AZ|}$. In our situation, we may as well start confusing segment lengths for complex norms and write $|AZ|=|Z|$ and $|AC|=3$, so $|AE|=\frac{9}{|Z|}$. To get $E$, we can normalize $Z$ and multiply by the length $|AE|$ to get $\frac{9}{|Z|}\frac{Z}{|Z|}=\frac{9Z}{|Z|^2}$.

The map $Z\mapsto \frac{9Z}{|Z|^2}$ is a Möbius transformation of the complex plane, and as such it will carry the given line onto a circle contained inside the circle radius $3$. It is also its own inverse function.

At the very worst, you can deduce the equation of the circle from the two points given above along with $(0,0)$.

But let's do better and deduce the equation! Let $m$ be the Möbius transformation above. Then if $(x,y)$ lie in the locus, $m(x+iy)$ lies on the given line, and therefore $m(x+iy)$ has to satisfy $4x-5iy=20$. In terms of $\Bbb R^2$, the transformation maps $(x,y)$ to $(\frac{9x}{x^2+y^2},\frac{9y}{x^2+y^2})$. To satisfy $4x-5y=20$, we have:

$$4(\frac{9x}{x^2+y^2})-5(\frac{9y}{x^2+y^2})=20$$

Rewriting:

$$4(9x)-5(9y)=20(x^2+y^2)\\ 36x-45y=20(x^2+y^2)\\ 0=20(x^2+y^2)-36x+45y$$

In standard form: $(x-\frac{9}{10})^2+(y+\frac{9}{8})^2 =\frac{3321}{1600}$