If $∆$ is the area of the triangle formed by the positive x-axis and the normal and tangent to the circle x² + y² = 4 at (1, $\sqrt{3}$) then $∆$ is equal to

• $\frac{\sqrt{3}}{2}$

• $\sqrt{3}$

• $2\sqrt{3}$

• 6

Given the circle C with the equation x² + y² - 2x + 10y - 38 = 0. Match the List I with the List II given below concerning C

	List I		List II
A	The equation of the polar of (4, 3)with respect to C	I	y + 5 = 0
B	The equation of the tangent at (9, - 5) on C	II	x = 1
C	The equation of  the normal at(- 7, - 5) on C	III	3x + 8y = 27
D	The equation of the diameter of  C passing through (1,3)	IV	x + y = 3
		V	x = 9

The correct answer is

For the circle C with the equation x² + y² - 16 - 12y + 64 = 0 match the List I with the List II given below.

	List I		List II
(i)	The equation of the polar of (- 5, 1) with respect to	(A)	y = 0
(ii)	The equation of the tangent at (8, 0) to C	(B)	y = 6
(iii)	The equation of the normal at (2, 6) to C	(C)	13x + 5y = 98
(iv)	The equation of the diameter of C through (8, 12)	(D)	13x + 5y = 98
		(E)	x = 6

The correct match is

The circle $4{\mathrm{x}}^2 + 4{\mathrm{y}}^2 - 12\mathrm{x} - 12\mathrm{y} + 9 = 0$

• Touches both the axes

• Touches the x-axis only

• Touches the y-axis only

• Does not touch the axes

If the length of the tangent from (h, k) to the circle x² + y² = 16 is twice the length of the tangent from the same point to the circle x² + y² + 2x + 2y = 0, then

• ${\mathrm{h}}^2 + {\mathrm{k}}^2 + 4\mathrm{h} + 4\mathrm{k} + 16 = 0$

• ${\mathrm{h}}^2 + {\mathrm{k}}^2 + 3\mathrm{h} + 3\mathrm{k} = 0$

• $3{\mathrm{h}}^2 + 3{\mathrm{k}}^2 + 8\mathrm{h} + 8\mathrm{k} + 16 = 0$

• $3{\mathrm{h}}^2 + 3{\mathrm{k}}^2 + 4\mathrm{h} + 4\mathrm{k} + 16 = 0$

366.

($\mathrm{\alpha }$, 0) and (b, 0) are centres of two circles belonging to a coaxial system of which y-axis is the radical axis. If radius of one of the circles is 'r', then the radius of the other circle is

• ${\left({\mathrm{r}}^2 + {\mathrm{b}}^2 + {\mathrm{a}}^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

• ${\left({\mathrm{r}}^2 + {\mathrm{b}}^2 - {\mathrm{a}}^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

• ${\left({\mathrm{r}}^2 + {\mathrm{b}}^2 - {\mathrm{a}}^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

• ${\left({\mathrm{r}}^2 + {\mathrm{b}}^2 + {\mathrm{a}}^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

If the circle x² + y² + 4x - 6y + c =0 bisects the circumference of the circle x² + y² - 6x + 4y - 12 = 0, then c is equal to

• 16

• 24

• - 42

• - 62

A circle of radius 4, drawn on a chord of the parabola y² = 8x as diameter, touches the axis of the parabola. Then, the slope of the chord is

• $\frac{1}{2}$

• $\frac{3}{4}$

• 1

• 2

The mid-point of a chord of the ellipse x² + 4y² - 2x + 20y = 0 is (2, - 4). The equation of the chord is

• x - 6y = 26

• x + 6y = 26

• 6x - y = 26

• 6x + y = 26

If the focus of the ellipse $\frac{{\mathrm{x}}^2}{25} + \frac{{\mathrm{y}}^2}{16} = 1$ and the hyperbola $\frac{{\mathrm{x}}^2}{4} - \frac{{\mathrm{y}}^2}{{\mathrm{b}}^2} = 1$ coincide, then b² =?

• 4

• 5

• 8

• 9