41.

The center of the circle (x - 2a) (x - 2b) + (y - 2c) (y – 2d) = 0

• (2a, 2c)

• (2b, 2d)

• (a + b, c + d)

• (a - b, c - d)

$$\begin{gathered} \left(\mathrm{x} - 2\mathrm{a}\right)\left(\mathrm{x} - 2\mathrm{b}\right) + \left(\mathrm{y} - 2\mathrm{c}\right)\left(\mathrm{y} - 2\mathrm{d}\right) = 0 \\ \Rightarrow {\mathrm{x}}^2 - 2\mathrm{ax} - 2\mathrm{bx} +\hspace{0.17em}4\mathrm{ab} +\hspace{0.17em}{\mathrm{y}}^2 - 2\mathrm{cy} - 2\mathrm{dy} + 4\mathrm{cd} = 0 \\ \Rightarrow {\mathrm{x}}^2 + {\mathrm{y}}^2 + 2\left( - \mathrm{a} - \mathrm{b}\right) + 2\left( - \mathrm{c} - \mathrm{d}\right)\mathrm{y} +\hspace{0.17em}4\mathrm{ab} +\hspace{0.17em}4\mathrm{cd} = 0 \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{equation} {\mathrm{x}}^2 + {\mathrm{y}}^2 +\hspace{0.17em}2\mathrm{gx} + 2\mathrm{fy} + \mathrm{c} = 0 \mathrm{represents} \mathrm{a} \mathrm{circle} \mathrm{having} \mathrm{centre} \mathrm{at} \left( - \mathrm{g}, - \mathrm{f}\right) \\ \mathrm{So} \mathrm{we} \mathrm{compare} {\mathrm{x}}^2 + {\mathrm{y}}^2 + 2\left( - \mathrm{a} - \mathrm{b}\right)\mathrm{x} + 2\left( - \mathrm{c} - \mathrm{d}\right)\mathrm{y} + 4\mathrm{ab} +\hspace{0.17em}4\mathrm{cd} = 0 \mathrm{with} \mathrm{standard} \mathrm{one} \\ \mathrm{Then} \mathrm{g} = \left( - \mathrm{a}, - \mathrm{b}\right) \mathrm{and} \mathrm{f} = \left( - \mathrm{c} - \mathrm{d}\right) \\ \mathrm{Then} \mathrm{centre} \mathrm{is} \left(\mathrm{a} +\hspace{0.17em}\mathrm{b}, \mathrm{c} + \mathrm{d}\right) \end{gathered}$$