Radical axis

Remarks

Expanding the two powers, the condition becomes $$2 \ip{P}{c_2 - c_1} = \norm{c_2}^2 - \norm{c_1}^2 - r_2^2 + r_1^2.$$ Since $c_2 - c_1 \ne 0$, this is a line perpendicular to $c_1 c_2$. For three circles with non-collinear centres, the three radical axes meet in one point, the radical centre.

Example: let $\Gamma_1$ be $x^2 + y^2 = 1$ and $\Gamma_2$ be $(x - 2)^2 + y^2 = 1$. Then $\pi_1 = x^2 + y^2 - 1$ and $\pi_2 = x^2 - 4x + 4 + y^2 - 1$, and they are equal exactly when $x = 1$.

Non-example: concentric circles fail the condition $c_1 \ne c_2$. For $c_1 = c_2$ and $r_1 \ne r_2$, the powers differ by the constant $r_1^2 - r_2^2 \ne 0$, so no point has equal powers.