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In coordinate geometry radical axis is defined as locus for which power is same with respect to two circles.

If we take two points as centre of two circles. Now without disturbing centre of circle if I increase radius of one circle will radical axis move away from it or get closer to it and vice versa.

Also if there is one big(radius) circle and another small(radius) circle. Is radical axis closer to centre of bigger(radius) circle or smaller(radius) circle.

Fawad
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1 Answers1

-1

Denote $A$ and $B$ the centers, and $R$ and $r$ the radii of the bigger and the smaller circle, respectively.

Second question

Radical axis $\mathcal{P}$ is orthogonal to $AB$, and is closer to the bigger circle. This follows immediately from $D^2-R^2=d^2-r^2,$ which is equivalent to $$D^2-d^2=R^2-r^2.\quad\quad\quad(1)$$ Here $D$ denotes the distance from a point on $\mathcal{P}$ to $A,$ and $d$ the distance from this same point to $B.$

First question

For the point $E$ common to $AB$ and $\mathcal{P},$ we have $D+d=|AB|,$ from where $$D^2-d^2=|AB|\,\left(|AB|-2d\right)$$ and $$D^2-d^2=|AB|\,\left(2D-|AB|\right)$$ Putting together with (1) we obtain $$|AB|\,\left(|AB|-2d\right)=R^2-r^2=|AB|\,\left(2D-|AB|\right).$$ If $A,B$ do not move and $r$ is constant, then increasing $R$ implies decreasing $d$ and increasing $D.$ That is, $\mathcal{P}$ gets closer to the CENTER of the smaller circle and moves away from the center of the bigger circle.

user376343
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  • How $D^2-d^2=R^2-r^2>0$ mean that the radical axis is closer to the circle with bigger radius? It means distance between the point on $\mathcal{P}$ and the center of the circle with radius $R$ is larger than $r$. It also means the perpendicular distance from radical axis to $A$ (center of the circle with bigger radius $R$) is larger than $B$ (center of the circle with smaller radius) from the fact that radical axis $\mathcal{P}$ is orthogonal to $AB$. – Nazmul Hasan Shipon Oct 17 '21 at 13:43
  • @NazmulHasanShipon Use Pythagorean theorem in triangles AMS and BMS, where A,B are the centers, M a point on the radical axis (different form S) and S the intersection of the radical axis and AB. To the author of the question, my answer was clear. Was it necessary to downvote simply because you do not see it? – user376343 Oct 18 '21 at 18:16
  • Well, this site contains generic questions which is supposed to be helpful for every audience who later come to visit that question. Now I am going to show you a user of this site with more than 200k reputation, who was suspended two months ago for low quality contribution like yours namely incomplete answers. Before commenting here, I have drawn the two right angled triangle that you're talking about. It shows that $AS$>$BS$. But what does that prove? How does it prove $AS-R < BS-r$? Please if you reply, answer it clearly. – Nazmul Hasan Shipon Oct 18 '21 at 20:26
  • @NazmulHasanShipon This is not the place for threatening. Do not forget that we all (who answer questions in MSE) are volunteers. By the way, the user you've mentioned is a blind person. He has my highest admiration. – user376343 Oct 19 '21 at 06:36
  • I am neither a moderator, nor a staff of Stack Exchange. So I think you shouldn't feel threatened by my comment. I just showed you a consequence of low quality contribution specifically unclear explanation or just like "do this, do that and you will get your answer" types answer. These things are best for comments. However, you didn't answer the question I was asking from my first comment. You may like to edit your original answer. Although previous 197 users didn't up vote your answer, a proper explanation will get you plenty of up votes in the future from future viewers and also from me. – Nazmul Hasan Shipon Oct 19 '21 at 13:29
  • And sorry I couldn't uncast my down vote to your happiness since it is locked. I can only uncast it if you edit your answer. – Nazmul Hasan Shipon Oct 19 '21 at 13:40