The points of intersection of the line $$ax + by = 0,(a \ne b)$$ and the circle $${x^2} + {y^2} - 2x = 0$$ are $$A(\alpha ,0)$$ and $$B(1,\beta )$$. The image of the circle with AB as a diameter in the line $$x + y + 2 = 0$$ is :
Let $$\mathrm{z_1=2+3i}$$ and $$\mathrm{z_2=3+4i}$$. The set $$\mathrm{S = \left\{ {z \in \mathbb{C}:{{\left| {z - {z_1}} \right|}^2} - {{\left| {z - {z_2}} \right|}^2} = {{\left| {{z_1} - {z_2}} \right|}^2}} \right\}}$$ represents a
Let $$\overrightarrow a $$, $$\overrightarrow b $$ and $$\overrightarrow c $$ be three non zero vectors such that $$\overrightarrow b $$ . $$\overrightarrow c $$ = 0 and $$\overrightarrow a \times (\overrightarrow b \times \overrightarrow c ) = {{\overrightarrow b - \overrightarrow c } \over 2}$$. If $$\overrightarrow d $$ be a vector such that $$\overrightarrow b \,.\,\overrightarrow d = \overrightarrow a \,.\,\overrightarrow b $$, then $$(\overrightarrow a \times \overrightarrow b )\,.\,(\overrightarrow c \times \overrightarrow d )$$ is equal to
Let S$$_1$$ and S$$_2$$ be respectively the sets of all $$a \in \mathbb{R} - \{ 0\} $$ for which the system of linear equations
$$ax + 2ay - 3az = 1$$
$$(2a + 1)x + (2a + 3)y + (a + 1)z = 2$$
$$(3a + 5)x + (a + 5)y + (a + 2)z = 3$$
has unique solution and infinitely many solutions. Then