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.
Best Answer
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
Please note that, if you had asked:
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:
$\tau $ isosurface /coordinates
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!