Explanation

(c)
The equation of tangent to the given circle $2x^{2} + 2y^{2} - 2x - 5y + 3 = 0$ at point (1, 1) is
$2x + 2y - (x + 1) - \frac{5}{2}(y + 1) + 3 = 0$
$\Rightarrow x - \frac{1}{2}y - \frac{1}{2} = 0$
$\Rightarrow 2x - y - 1 = 0$
$\Rightarrow y = 2x - 1$
Slope of tangent=2, therefore slope of normal $= - \frac{1}{2}$
Hence, equation of normal at point (1, 1) and having slope $\left( - \frac{1}{2} \right)$ is
$y - 1 = - \frac{1}{2}(x - 1)$
$\Rightarrow 2y - 2 = - x + 1$
$\Rightarrow x + 2y = 3$