Answer
The families of the given curves intersect at $ x=\frac{2c^2k^2} {k(c^2+k^2)} $ and $y=\frac{2ck^2} {c^2+k^2} $ with slopes $\frac{k^2-c^2} {2ck} $ and $-\frac{2ck} {k^2-c^2} $.
The slope are negative reciprocal of each other.
Hence families of the curves are orthogonal trajectories of each other.
Work Step by Step
$ x^2+(y-c) ^2=c^2$
$x^2+y^2-2yc+c^2=c^2$
$x^2+y^2-2yc=0 $ .............. eq (1)
$(x-k) ^2+y^2=k^2$
$x^2+k^2-2kx+y^2=k^2$
$x^2+y^2 -2kx=0$ .................. eq (2)
Subtracting equation (2) from equation (1)
$kx=cy$ Imply that $x=\frac{c}{k} y$ .............. eq (2')
Putting equation (2') in equation (1)
$ (\frac{c}{k} y) ^2+y^2-2yc=0$
$\frac{c^2} {k^2} y^2+y^2-2yc=0$
$ (\frac{c^2} {k^2} +1) y^2=2yc$
$(\frac{c^2+k^2} {k^2}) y^2=2yc$
$(\frac{c^2+k^2} {k^2}) y=2c$
$y=\frac{2ck^2} {c^2+k^2} $ ............. eq (3)
Putting equation (3) in equation (2')
$x=\frac{c} {k}\frac{2ck^2} {c^2+k^2} $
$ x=\frac{2c^2k^2} {k(c^2+k^2)} $ ................... eq (4)
Differentiating equation (2) with respect to x
$\frac{d(x^2+y^2-2kx)} {dx}=\frac{d(0)}{dx}$
$ \frac{dx^2} {dx}+\frac{d(y^2)}{dx}-\frac{d(2kx)}{dx}=0 $
$2x+\frac{d(y^2)} {dy}\frac{dy}{dx}-2k\frac{dx}{dx} =0$
$2x+2y\frac{dy}{dx}-2k=0$
$2y\frac{dy}{dx}=2k-2x=2(k-x) $
$\frac{dy}{dx}=\frac{k-x} {y}$ ................. eq (5)
Putting equation (3) and equation (4) in equation (5)
$\frac{dy}{dx}=\frac {k- \frac{2c^2k^2} {k(c^2+k^2)}} {\frac{2ck^2} {c^2+k^2}} $
$\frac{dy}{dx}=\frac {[k- \frac{2c^2k^2} {k(c^2+k^2)}]} {[\frac{2ck^2} {c^2+k^2}]} \frac{k(c^2+k^2)} {k(c^2+k^2)} $
$\frac{dy}{dx}=\frac {[k- \frac{2c^2k^2} {k(c^2+k^2)}]} {[\frac{2ck^2} {c^2+k^2}]} \frac{k(c^2+k^2)} {k(c^2+k^2)} =\frac{k^2(c^2+k^2)-2c^2k^2}{2ck^2k}$
$\frac{dy}{dx} =\frac{k^2(c^2+k^2)-2c^2k^2}{2ck^2k} =\frac{k^2c^2+k^4-2c^2k^2}{2ck^2k} =\frac{k^4-c^2k^2} {2ck^2 k} =\frac{k^2(k^2-c^2)}{2ck^2k} =\frac{k^2-c^2}{2ck}=m_1$............. eq (6)
Where $m_1$ is the slope of the tangent to the curve
Differentiating equation (1) with respect to x
$\frac{d(x^2+y^2-2yc)}{dx}=\frac {d (0)} {dx}$
$ \frac{d(x^2)} {dx}+\frac{d(y^2)} {dy}\frac{dy}{dx}-2c\frac{dy}{dx}=0$
$2x+2y\frac{dy}{dx}-2c\frac{dy}{dx}=0$
$2(y-c)\frac{dy}{dx}=-2x$
$\frac{dy}{dx}=-\frac{x}{y-c} = $ ................ eq (7)
Putting equation (3) and equation (4) in equation (7)
$\frac{dy}{dx}=-\frac{\frac{2c^2k^2} {k(c^2+k^2)}} {(\frac{2ck^2} {c^2+k^2}) -c} $
$\frac{dy}{dx}=-[\frac{\frac{2c^2k^2} {k(c^2+k^2)}} {(\frac{2ck^2} {c^2+k^2}) -c}] \frac{k(c^2+k^2)}{k(c^2+k^2)} =-\frac{2c^2k^2}{2ck^3-ck(c^2+k^2)} =-\frac{2ck}{k^2-c^2}=m_2$ .............. eq (8)
Where $m_2$ is the slope of the tangent to the curve
Multiplying equation (6) and equation (8)
Now
$m_1m_2=\frac{k^2-c^2}{2ck}(-\frac{2ck}{k^2-c^2}) =-1$
Since tangents lines are mutually perpendicular at their point of intersection, families of the given curves are orthogonal trajectories of one another
