If the equation of an ellipse is 3x² + 2y² + 6x - 8y + 5 = 0, then which of the following are true?

• e = $\frac{1}{\sqrt{3}}$

• centre is (- 1, 2)

• foci are (- 1, 1) are (- 1, 3)

• All of the above

The equation of the common tangents to the two hyperbolas $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2} - \frac{{\mathrm{y}}^2}{{\mathrm{b}}^2} = 1 \mathrm{and} \frac{{\mathrm{y}}^2}{{\mathrm{a}}^2} - \frac{{\mathrm{x}}^2}{{\mathrm{b}}^2} = 1, \mathrm{are}$

• $\mathrm{y} = \pm \mathrm{x} \pm \sqrt{{\mathrm{b}}^2 - {\mathrm{a}}^2}$

• $\mathrm{y} = \pm \mathrm{x} \pm \sqrt{{\mathrm{a}}^2 - {\mathrm{b}}^2}$

• $\mathrm{y} = \pm \mathrm{x} \pm \sqrt{{\mathrm{a}}^2 + {\mathrm{b}}^2}$

• $\mathrm{y} = \pm \mathrm{x} \pm \left({\mathrm{a}}^2 - {\mathrm{b}}^2\right)$

The equation of sphere concentric with the sphere x² + y² + z² - 4x - 6y - 8z - 5 = 0 and which passes through the origin, is

• x² + y² + z² - 4x - 6y - 8z = 0

• x² + y² + z² - 6y - 8z = 0

• x² + y² + z² = 0

• x² + y² + z² - 4x - 6y - 8z - 6 = 0

The equation ${\mathrm{\lambda x}}^2 + 4\mathrm{xy} + {\mathrm{y}}^2 + \mathrm{\lambda x} + 3\mathrm{y} + 2 = 0$ represents parabola, if $\lambda$ is

• 0

• 1

• 2

• 4

385.

If two circles 2x² + 2y² - 3x + 6y + k = 0 and x² + y² - 4x + 10y + 16 = 0 cut orthagobally then, value of k is

• 41

• 14

• 4

• 1

The locus of z satisfying the inequality $\left|\frac{\mathrm{z} +\hspace{0.17em}2\mathrm{i}}{2\mathrm{z} +\hspace{0.17em}\mathrm{i}}\right| < 1$, where z = x + iy, is

• x² + y² < 1

• x² - y² < 1

• x² + y² > 1

• 2x² + 3y² < 1

The area (in square unit) of the circle which touches the lines 4x + 3y = 15 and 4x + 3y = 5 is

• 4$\mathrm{\pi }$

• $3\mathrm{\pi }$

• $2\mathrm{\pi }$

• $\mathrm{\pi }$

The equations of the circle which pass through the origin and makes intercepts of lengths 4 and 8 on the x and y -axes respectively are

• x² + y² ± 4x ± 8y = 0

• x² + y² ± 2x ± 4y = 0

• x² + y² ± 8x ± 16y = 0

• x² + y² ± x ± y = 0

The point (3 -4) lies on both the circles x² + y² - 2x + 8y + 13 = 0 and x² + y² - 4x + 6y + 11 = 0. Then, the angle between the circles is

• $60°$

• ${\tan }^{-1}\left(\frac{1}{2}\right)$

• ${\tan }^{-1}\left(\frac{3}{5}\right)$

• 135$°$

The equation of the circle which passes through the origin and cuts orthogonally each of the circles x² + y² - 6x + 8 = 0 and x² + y² - 2x - 2y = 7 is

• 3x² + 3y² - 8x - 13y = 0

• 3x² + 3y² - 8x + 29y = 0

• 3x² + 3y² + 8x + 29y = 0

• 3x² + 3y² - 8x - 29y = 0