This the question in my text-book:
The tangent to $x^2 + y^2 = a^2$ having inclination $\alpha$ and $\beta$ intersect at $P$. If $\cot\alpha$ + $\cot\beta = 0$, then the locus of $P$ is...
I really don't know how to approach this question so any help will be appreciable.
On
Starting with $\cot \alpha+\cot \beta=0$, we have the following:
$$\sin \alpha\cos \beta+\sin \beta\cos \alpha=0$$
$$\implies \sin(\alpha+\beta)=0\implies \alpha+\beta=0+n\pi$$
Then we have that $\alpha=n\pi-\beta$. From this, we can conclude that one of the $x,y$ coordinates of $P$ is $0$. Specifically, the angle of incidence is mirrored about the $0+n\pi$ angle, and the circle is centered at the origin (as specified by the equation of it), and thus the tangent lines are mirrored about the $x$ or $y$ axis as well.
For the non-zero coordinate, consider the triangle whose hypotenuse is that coordinate, and which uses $a$ as the length of one of its legs. Note that the third leg of this triangle is the segment along one of the tangent lines described in the question stretching from the tangent point with the circle to the axis where it meets in the angle $\alpha$. The angle $\alpha$ is complementary to the angle adjacent to the $a$ leg in this triangle, so we would use $a\csc\alpha$ as the value of the length of the hypotenuse, which means that our point $P$ occurs at
$$P=({\pm a\over\sin\alpha},0)$$
or
$$P=(0,{\pm a\over\sin\alpha})$$
On
HINT:
As $\cot\alpha+\cot\beta=0$ we have $\tan\alpha=-\tan\beta=\tan(-\beta)$
$\implies \alpha=n\pi-\beta$ where $n$ is any integer
So, here $\alpha=-\beta$ or $\pi-\beta$
From the Article $148,150$ of The elements of coordinate geometry by Loney,
the equation of the tangent at $(x_1,y_1)$ of the Circle $x^2+y^2=a^2$ is $xx_1+yy_1=a^2$
Now, the Parametric Equation of the circle is $x=a\cos\theta,y=a\sin\theta$
So, the equation of the tangent becomes $x\cos\theta+y\sin\theta=a\iff y=-x\cot\theta+a\csc\theta$ whose gradient is $-\cot\theta=\cot(-\theta)$ or $\cot(\pi-\theta)$
Replace $\theta$ with $\alpha,\beta$ to find two simultaneous equation in $\sin\beta,\cos\beta$
Eliminate $\beta$ using $\sin^2\beta+\cos^2\beta=1$
On
We use the following notations:-
The center of the circle is O.
P is at (X, Y).
The points of contacts are A and B.
Some of the answers confirmed that the tangents cut each other at right angles. We continue from that point on and finish the rest (which is then just a one line proof.)
From the figure, OAPB is a square. Thus,
$X^2 + Y^2 = OP^2 = AB^2 = 2a^2$
And hence the locus of P is $x^2 + y^2 = 2a^2$
The answer which was checked and upvoted was wrong. Hope this one is better.
Let T be the intersection point of the 2 lines.
If cot$\alpha$ +cot$\beta$ = 0 the same relationship holds for the tan which is 1/cot. So tan$\alpha$ = -tan$\beta$ and $\alpha = -\beta \pm n\pi$. Let us assume that $\alpha = -\beta$ and that line 1 intersects the x-axis at point (b,0). Line 2 is the same line with negative slope, so it has to intersect the x-axis at the point (-b,0).
So the triangle with vertices (-b,0),(b,0) and T is isoceles. If we knew what b is, we would be done.
What we do know is that line 1 is tangent to the circle at a point p = $a(cos\phi,sin\phi)$ where $\phi = \frac{\pi}{2} - \theta$. This is because the radius from (0,0) to $a(cos\phi,sin\phi)$ is perpendicular to line 1; so the triangle with vertices (0,0),(0,b) and $a(cos\phi,sin\phi)$ is a right triangle. That doesn't leave any choice for $\phi$.
The distance from p to the y-axis is then $acos\phi$. The height h of the isoceles triangle with vertices $a(cos\phi,sin\phi)$,$(-acos\phi,asin\phi)$, and T is h = $a(cos\phi)(sin\theta)$ (since $\theta$ is the base angle for that triangle).
So the height of the entire triangle with vertices at (-b,0), (b,0) and T is $asin\phi + a(cos\phi)(sin\theta)$ which you can simplify down to $asin\theta(1+sin\theta)$.