All the circles having centres on a fixed line and passing thorugh a fixed point outside the fixed line; also passes through another fixed point

Prove that all the circles having centres on a fixed line and passing thorugh a fixed point (not lying on the fixed line) also passes through another fixed point.

My attempt- Let the fixed line be $y=mx+c$ let the fixed point be $P:(h,k)$. Now equation of any circle with center in the line $y=mx+c$ will look like

$(x-a)^2+(y-b)^2=r^2$ with the conditions $b=ma+c$ and $(h-a)^2+(k-b)^2=r^2$ .I am stuck here.

Geometric solutions are welcome , but I am looking for an analytical solution.

• Try geometrically. Draw a straight line and a point, and draw 5 different circles having center on the line. Can you see any pattern?
– 5xum
Oct 24 '16 at 10:18
• I have a feeling it is the image of the fixed point along the fixed line. Oct 24 '16 at 10:20
• You mean the mirror image?
– 5xum
Oct 24 '16 at 10:21
• Yes, taking the fixed line as the mirror Oct 24 '16 at 10:22
• Find the location of the other fixed point. Then you can easily show that the distance from one fixed point to a point on the line is always the same as the distance from that point on the line to the other fixed point. Oct 24 '16 at 10:52

Although $$(R,h)$$ are variable, difference of their squared distances $$(R^2-h^2)=$$ should be constant $$a^2$$ as shown in the picture following.

So with every fixed point A1 there is a corresponding fixed point A2 mirrored about $$x-$$ axis.

$$(x-h)^2+ y^2 = h^2 + a^2 \,or \quad x^2-2xh +y^2 = a^2 \tag1$$

where $$h$$ is variable point $$B$$ is moving on x-axis and $$a$$ is constant. The above circle with $$B$$ as center is now drawn in black.

EDIT1:

Points $$(A1,A2),(OA1=OA2=a)$$ are singular points, meaning all circles pass through them. They are obtained as singular solutions of DE derived by partial differentiation with respect

to $$h$$ and eliminating it from 1)

$$x^2-2 x\,h +y^2=a^2\, \rightarrow x =0 \rightarrow y= \pm a \tag2$$

The distance $$OA$$ in fact serves as one among two bipolar $$\sigma$$ isosurface /coordinates.

EDIT2

Prove that all the circles having centres on a fixed line and whose tangents pass through a fixed point , lying on that fixed line must have another tangent passing through this fixed point,

then the constant in this case gets its sign simply changed to $$(h^2-R^2)= a^2$$ when we are now switching to their orthogonal trajectories:

which have equi-tangent constant lengths $$=a$$ circles of equation

$$x^2+y^2-2 x h -R^2= a^2 \tag3$$

I have included this as your preference is for the analytical approach. But do not lose the geometrical connection! The given line $\ell$ is a symmetry axis for each of these circles. Therefore all of them will go through the point $P'$ obtained from reflecting $P$ in $\ell$.

Set up an orthonormal coordinate system so that the fixed line is the $$x$$-axis and the point $$P$$ lies on the $$y$$-axis. That is, the equation of the line is $$y = 0$$ and $$P = (0,y_P).$$

This is always possible to do. Even if the line and point have already been given in terms of some other coordinates, you can transform the coordinates to the desired form.

Now the equation of the radius of the circle with center $$(0,x)$$ is $$r^2 = x^2 + y_P^2.$$ But $$x^2 + y_P^2 = x^2 + (-y_P)^2,$$ that is, if $$(0,y_P)$$ lies on the circle then so does $$(0,-y_P)$$. But we are given that $$(0,y_P)$$ lies on every circle, and therefore so does $$(0,-y_P)$$.