So, I was doing some analytical geometry, circles to be precise. There, the concept of polar was introduced as: "If through a point P (within or without the circle) there be drawn any straight line to meet the circle in Q and R, the locus of the point of intersection of tangents at Q and R is called the polar of P; also P is called the pole of the polar."
Now, this locus comes out to be of the form of a straight line which passes through the circle if P lies outside the circumference of the given circle. My doubt is, if the polar is a locus of points of intersection of tangents of a circle, how can it pass through the inside of the circle since that would mean that there are intersection points of two tangents of circle lying inside it, which is downright impossible. Are there "imaginary" points of contact to be considered.
P.S.: The book I refer to is The Elements of Coordinate Geometry by S.L. Loney, it hasn't left me in doubts yet.
Polar of a circle.
analytic geometrycirclesgeometry
Related Solutions
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).
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:
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: $(x-\frac{9}{10})^2+(y+\frac{9}{8})^2 =\frac{3321}{1600}$
Best Answer
$P$ inside the circle:
$P$ outside the circle:
Since $\triangle PUO\sim\triangle STO$ (two right triangles sharing the acute angle at $O$) $$ \overline{PO}\,\,\overline{TO}=\overline{UO}\,\,\overline{SO}\tag1 $$ Since $\triangle QUO\sim\triangle SQO$ (two right triangles sharing the acute angle at $O$) $$ \overline{UO}\,\,\overline{SO}=\overline{QO}\,\,\overline{QO}\tag2 $$ Therefore, $$ \overline{PO}\,\,\overline{TO}=\overline{QO}^{\large\,2}\tag3 $$ When $P$ is outside the circle, the intersection of the tangents is outside the circle, but it lies on the line perpendicular to $\overline{PO}$ at the point $T$ satisfying $(3)$.