It’s seen that the given bends are condition of two circles.
\[2x\text{ }=\text{ }y2\text{ }\ldots \text{ }..\text{ }\left( 1 \right)\] and
\[2xy\text{ }=\text{ }k\text{ }\ldots \text{ }..\text{ }\left( 2 \right)\]
We realize that, two circles converge symmetrically if the point between the digressions attracted to the two circles at the place of their convergence is 90o.
Presently, separating conditions (1) and (2) w.r.t. t, we get
Presently, settling conditions (1) and (2), we have
\[y\text{ }=\text{ }k/2x\text{ }\left[ From\text{ }\left( 2 \right) \right]\]
Placing the worth of y in condition (1),
\[2x\text{ }=\text{ }\left( k/2x \right)2\Rightarrow 2x\text{ }=\text{ }k2/4×2\]
\[8×3\text{ }=\text{ }k2\]
\[8\left( 1 \right)3\text{ }=\text{ }k2\]
\[k2\text{ }=\text{ }8\]
Subsequently, the necessary condition is \[k2\text{ }=\text{ }8.\]