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(B) The angle subtended by the chord at the center of the circle $x^{2} + y^{2} = 4$ is $2 	imes 45^{\circ} = 90^{\circ}$.Let the midpoint of the cho

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$x^{2} + y^{2} = 2$

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(B) The angle subtended by the chord at the center of the circle $x^{2} + y^{2} = 4$ is $2 	imes 45^{\circ} = 90^{\circ}$.Let the midpoint of the chord $AB$ be $(h, k)$. The equation of the chord $AB$ is given by $T = S_1$,which is:$hx + ky - 4 = h^{2} + k^{2} - 4$$hx + ky = h^{2} + k^{2}$$\Rightarrow \frac{hx + ky}{h^{2} + k^{2}} = 1$Using this line to homogenize the circle equation $x^{2} + y^{2} - 4 = 0$,we get:$x^{2} + y^{2} - 4 \left( \frac{hx + ky}{h^{2} + k^{2}} ight)^{2} = 0$This represents the combined equation of lines $OA$ and $OB$. Since the angle at the center is $90^{\circ}$,the lines $OA$ and $OB$ are perpendicular.For perpendicular lines,the sum of the coefficients of $x^{2}$ and $y^{2}$ must be zero:$(1 - \frac{4h^{2}}{(h^{2} + k^{2})^{2}}) + (1 - \frac{4k^{2}}{(h^{2} + k^{2})^{2}}) = 0$$2 - \frac{4(h^{2} + k^{2})}{(h^{2} + k^{2})^{2}} = 0$$2 - \frac{4}{h^{2} + k^{2}} = 0$$h^{2} + k^{2} = 2$Thus,the locus is $x^{2} + y^{2} = 2$.

The locus of the midpoint of a chord of the circle $x^{2} + y^{2} = 4$ which subtends an angle of $45^{\circ}$ at the major arc of the circle is:

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