Explanation

(b)
Given, circle cuts the parabola
$\therefore x^{2} + \left( \frac{x^{2}}{4a} \right)^{2} + 2gx + 2f\left( \frac{x^{2}}{4a} \right) + c = 0$
$\Rightarrow x^{4} + {16a}^{2}x^{2} + {8afx}^{2} + {32gxa}^{2} + {16a}^{2}c = 0$
$\sum_{}^{}{x_{i} = 0}$ …(i)
$\sum_{}^{}{x_{1}x_{2} = {16a}^{2} + 8af}$ …(ii)
Now, $\sum_{}^{}{y_{i} = \frac{1}{4a}\sum_{}^{}x_{i}^{2}}$
$= \frac{1}{4a}\lbrack\left( x_{1} + x_{2} + x_{3} + x_{4} \right)^{2} - 2\sum_{}^{}{x_{1}x_{2}\rbrack}$
$= - \frac{1}{2a}\left( {16a}^{2} + 8af \right) = - 4(f + 2a)$