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

192.

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

The number of normals drawn to the parabola y² = 4x from the point (1, 0) is

• 0

• 1

• 2

• 3

If the circle x² + y² = a intersects the hyperbola xy = c² in four points (x_i, y_i), for i = 1, 2, 3 and 4, then y₁ + y₂ + y₃ + y₄ equals

• 0

• c

• a

• c⁴

The mid point of the chord 4x - 3y = 5 of the hyperbola 2x² - 3y² = 12 is

• $\left(0, - \frac{5}{3}\right)$

• (2, 1)

• $\left(\frac{5}{4}, 0\right)$

• $\left(\frac{11}{4}, 2\right)$