$$\begin{gathered} \mathrm{The} \mathrm{point} \mathrm{of} \mathrm{contact} \mathrm{of} \mathrm{the} \mathrm{circles} \\ {\mathrm{x}}^2 + {\mathrm{y}}^2 + 2\mathrm{x} + 2\mathrm{y} + 1 = 0 \mathrm{and} \\ {\mathrm{x}}^2 + {\mathrm{y}}^2 - 2\mathrm{x} +\hspace{0.17em}2\mathrm{y} + 1 = 0 \mathrm{is} \end{gathered}$$

• (0 , 1)

• (0, - 1)

• (1, 0)

• (- 1, 0)

The polar equation of the line perpendicular to the line $\sin \left(\mathrm{\theta }\right) - \cos \left(\mathrm{\theta }\right) = \frac{1}{\mathrm{r}}$ and passing through the point $\left(2, \frac{\mathrm{\pi }}{6}\right) \mathrm{is}$

• $\sin \left(\mathrm{\theta }\right) + \cos \left(\mathrm{\theta }\right) = \frac{\sqrt{3} + 1}{\mathrm{r}}$

• $\sin \left(\mathrm{\theta }\right) - \cos \left(\mathrm{\theta }\right) = \frac{\sqrt{3} + 1}{\mathrm{r}}$

• $\sin \left(\mathrm{\theta }\right) + \cos \left(\mathrm{\theta }\right) = \frac{\sqrt{3} - 1}{\mathrm{r}}$

• $\cos \left(\mathrm{\theta }\right) - \sin \left(\mathrm{\theta }\right) = \frac{\sqrt{3}}{\mathrm{r}}$

The equation of a straight line passing through the point (1, 2) and inclined at 45° to the line y = x + 1 is

• 5x + y = 7

• 3x + y = 5

• x + y = 3

• x - y + 1 = 0

$$\begin{gathered} \mathrm{The} \mathrm{distance} \mathrm{between} \mathrm{the} \mathrm{parallel} \mathrm{lines} \mathrm{given} \mathrm{by} \\ {\left(\mathrm{x} + 7\mathrm{y}\right)}^2 + 4\sqrt{2}\left(\mathrm{x} +\hspace{0.17em}7\mathrm{y}\right) - 42 = 0 \mathrm{is} \end{gathered}$$

• $\frac{4}{5}$

• $4\sqrt{2}$

• 2

• $10\sqrt{2}$

175.

A straight line is equally inclined to all the three coordinate axes. Then, an angle made by the line with the y-axis is

• ${\cos }^{-1}\left(\frac{1}{3}\right)$

• ${\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$

• ${\cos }^{-1}\left(\frac{2}{\sqrt{3}}\right)$

• $\frac{\mathrm{\pi }}{4}$

If p and q are the perpendicular distances from the origin to the straight lines xsec$\left(\mathrm{\theta }\right)$ - ycosec($\left(\mathrm{\theta )}\right)$ = $\mathrm{\alpha }$ and xcos($\left(\mathrm{\theta }\right)$) + ysin($\left(\mathrm{\theta }\right)$) = $\mathrm{\alpha }$cos(2$\left(\mathrm{\theta )}\right)$, then

• 4p² + q² = a²

• p² + q² = a²

• p² + 2q² = a²

• 4p² + q² = 2a²

If 2x + 3y =5 is the perpendicular bisector of the line segment joining the points A (1, 1/3) and B, then B is equal to

• $\left(\frac{21}{13}, \frac{49}{39}\right)$

• $\left(\frac{17}{13}, \frac{31}{39}\right)$

• $\left(\frac{7}{13}, \frac{49}{39}\right)$

• $\left(\frac{21}{13}, \frac{31}{39}\right)$

If the points (1, 2) and (3, 4) lie on the same side of the straight line 3x - 5y + a = 0, then a lies in the set

• [7, 11]

• R - [7, 11]

• $\left(7, \infty \right)$

• $\left(- \infty , 11\right)$

If the equation ${\mathrm{ax}}^2 + 2\mathrm{hxy} + {\mathrm{by}}^2 +\hspace{0.17em}2\mathrm{gx} + 2\mathrm{fy} + \mathrm{c} = 0$represents a pair of, straight lines, then the square of the distance of their point of intersection from the origin is

• $\frac{\mathrm{c}\left(\mathrm{a} + \mathrm{b}\right) - {\mathrm{af}}^2 - {\mathrm{bg}}^2}{\mathrm{ab} - {\mathrm{h}}^2}$

• $\frac{\mathrm{c}\left(\mathrm{a} + \mathrm{b}\right) + {\mathrm{f}}^2 + {\mathrm{g}}^2}{\mathrm{ab} - {\mathrm{h}}^2}$

• $\frac{\mathrm{c}\left(\mathrm{a} + \mathrm{b}\right) - {\mathrm{f}}^2 - {\mathrm{g}}^2}{\mathrm{ab} - {\mathrm{h}}^2}$

• $\frac{\mathrm{c}\left(\mathrm{a} + \mathrm{b}\right) - {\mathrm{f}}^2 - {\mathrm{g}}^2}{{\left(\mathrm{ab} - {\mathrm{h}}^2\right)}^2}$

The equation of a straight line; perpendicular to 8x - 4y = 6 and forming a triangle of area 6sq. units with coordinate axes, is

• x - 2y = 6

• 4x + 3y = 12

• 4x + 3y + 24 = 0

• 3x + 4y = 12