0

enter image description here

I'm unable to understand this concept. I mean by using some manipulation we got the equation of a circle but I can't understand why this equation represents the intersection of these two lines. I need some insight on this method (exactly what's going on here) or maybe some another approach.

Navdeep Singh
  • 333
  • If $(x,y)$ satisfies $f(x,y) = a$ and $g(x,y) = b$. Then first, $(x,y)$ satisfies both $f^2(x,y) =a^2$ and $g^2(x,y) = b^2$. Finally, $(x,y)$ satisfies $f^2(x,y) + g^2(x,y) = a^2 + b^2$. Note that some convenient cancellation occurs when you add the two equations. Strictly, the locus of points satisfying the sum of squares could contain extraneous solutions (squaring will do that), but such does not happen in this case (as can be verified by plugging every point satisfying the sum of squares equation back into the original equations). – Eric Towers Nov 08 '21 at 05:34

1 Answers1

2

What's happening here is that $a$ and $b$ are chosen ahead of time to be some pair of values. Then the locus we are trying to find is defined by considering where the two lines intersect for each possible value of $\theta$. So, for example, when $\theta = 0$, the lines intersect at $(a, -b)$; when $\theta = \frac{\pi}{4}$, the lines intersect at $(\frac{a+b}{\sqrt{2}}, \frac{a-b}{\sqrt{2}}$), and so forth.

Since we're interested in the coverage of these intersection points over all possible values of $\theta$, we want to find some way to describe a generic point in the locus independently of $\theta$, hence why we eliminate $\theta$ from the equation. When we do so, we find that the points all satisfy the relation $x^2 + y^2 = a^2 + b^2$, which is the equation of a circle. Now technically there is one step missing - while we know that all of the points lie on the circle, we don't know if every point in the circle is represented in the locus. However, hopefully with a bit of algebra you can convince yourself that this is indeed the case.

ConMan
  • 24,300