Solving Orthogonal Circles Problem

[zorro]

Homework Statement

A member of the family of the circles that cuts all the members of the family of circles
x^2 + y^2 + 2gx + c=0 orthogonally, where c is a constant and g is a parameter is?

Homework Equations

The Attempt at a Solution

Let the equation of the required circle be x^2 + y^2 + 2g1x + 2f1y + c1 =0
This circle cuts the given circle orthogonally.
Therefore 2g1g + 2f1f = c1 + c
IDK how to proceed after this.

 

[HallsofIvy]

Are you allowed to assume that the curves that cut the given family of circles (not a single circle) is a circle?

Assuming you are, then you need to look at the derivatives to find the slopes of the tangent lines or, perhaps better, look at the tangent lines at points of intersection.

I think I would be inclined to write the first family of circles as
$$x^2 + 2gx+ y^2= -c$$
or
$$(x+ g)^2+ y^2= g^2- c$$
so this is a circle with center at (-g, 0) and radius $r= \sqrt{g^2- c}$

Parametric equations for such a circle are $x= r cos(\theta)- g$, $y= r sin(\theta)$ and the tangent vector at any point is given by $-r sin(\theta)\vec{i}+ r cos(\theta)\vec{j}$.

Any "orthogonal circle" can be written as $(x- a)^2+ (y- b)^2= R^2$ or in parametric equations, $x= R cos(\phi)+ a$, $y= R sin(\phi)+ b$ and has tangent vector at each point $-R sin(\phi)\vec{i}+ R cos(\phi)\vec{j}$.

To be orthogonal, at each point at which $r cos(\theta)- g= R cos(\phi)+ a$ and $r sin(\theta)= R sin(\phi)+ b$, we must have the dot product of the tangent vectors equal to 0: $Rr sin(\theta) sin(\phi)+ rR cos(\theta)cos(\phi)= 0$ which implies $sin(\theta)sin(\phi)+ cos(\theta)cos(\phi)=$$cos(\theta)cos(-\phi)- sin(\theta)sin(-\phi)= cos(\theta- \phi)= 0$.

If you are not allowed to assume from the start that the orthogonal family is a family of circles, then this becomes a problem in differential equations.

First divide the equation $x^2+ y^2+ 2gx+ c= 0$ by x so that the parameter "g" will be isolated without being multiplied by x:
$$x+ \frac{y^2}{x}+ 2g+ \frac{c}{x}= 0$$.

Differentiate with respect to x:
$$1+ \frac{2yy'- y^2}{x^2}- \frac{c}{x^2}= 0$$
no longer has the parameter "g" and so is true for every member of the family. y' here is the slope at each point for each member of the family of circles. If Y(x) is curve orthogonal to every member of that family, then Y'= -1/y' or y'= -1/Y'.

So members of the orthogonal family must satisfy
$$1+ \frac{2Y(-1/Y')- Y^2}{x^2}- \frac{c}{x^2}= 0$$

 

[zorro]

Thanks for your reply.
It is given in the question that the curves which cut the given family of circles orthogonally are circles itself.

A different approach has been adapted in my book-

'x^2 + y^2 + 2gx +c=0 represents the family of circles with centre on x-axis and radical axis as y-axis. So from the point (0,β), lying on the radical axis, tangents of equal length to all members of given family can be drawn.

Therefore circle of the form x^2 + (y-β)^2 = β^2 + c cuts all the circles of given family orthogonally.'

How is the radius of the above circle derived?

 

[HallsofIvy]

Actually, I think it shouldn't be what you have. The radius of that circle should be $\sqrt{\beta^2-c}$, not $\sqrt{\beta^2+ c}$ as you have.

Since these circles are orthogonal to all of the circles in the original family, they are, in particular, orthogonal to the circle with g= 0, $x^2+ y^2+ c= 0$ which is a circle with center at the origin and radius $\sqrt{-c}$ (of course, c must be negative). Let the radius of that circle be r and the radius of the circle orthogonal to it be R. The line from $(0, \beta)$ to a point, (x, y) on that circle, being a radius of the orthogonal circle has length R and is perpendicular to the radius of that circle, the line from (0, 0) to (x, y), which has length r. They form a right triangle with hypotenuse the line from (0, 0) to $(0, \beta)$ which has length $\beta$.

By the Pythagorean theorem, $R^2+ r^2= \beta^2$ so that $R^2= \beta^2- r^2$. The original circle, as I said, has length $\sqrt{-c}$ and so $r^2= -c$. We have $R^2= \beta^2+ c$, not $R^2= \beta^2+ c$.

Of course, the whole problem would make more sense (c would not have to be negative) if the equation were $x^2+ y^2+ 2gx- c= 0$ or $x^2+ y^2+ 2gx= c$ rather than $x^2+ y^2+ 2gx+ c= 0$. In that cases $R^2= \beta^2- c$ would be correct.

 

[zorro]

Thanks a lot for your detailed reply.
Actually $\sqrt{\beta^2-c}$ $\sqrt{\beta^2+c}$
would not matter as 'c' is a constant (given). So -c and c both mean the same.