The centre and radius of the spehere `x^2+y^2+z^2=3x-4z+1=0` are (A) `(-3/2, 0, -2),sqrt(21)/2` (B) `(-3/2, 0, 2),sqrt(21)/2` (C) `(-3/2, 0, -2),sqrt(21)/2` (D) `(-3/2, 2, 0),21/2`

The centre of a circle passing through (0,0), (1,0) and touching the CircIe x^2+y^2=9 is a. (1/2,sqrt2) b. (1/2,3/sqrt2) c. (3/2,1/sqrt2) d. (1/2,-1/sqrt2) .

The true solution set of inequality (log)_((x+1))(x^2-4)>1 is equal to (a) (2, ∞) (b) (2, (1+sqrt(21))/2) (c) ((1-sqrt(21))/2,(1+sqrt(21))/2) (d) ( (1+sqrt(21))/2 , ∞)

The equation of a circle of radius 1 touching the circles x^2+y^2-2|x|=0 is (a) x^2+y^2+2sqrt(2)x+1=0 (b) x^2+y^2-2sqrt(3)y+2=0 (c) x^2+y^2+2sqrt(3)y+2=0 (d) x^2+y^2-2sqrt(2)+1=0

Radius of the circle x^2+y^2+2x costheta+2y sin theta-8=0 is 1 2. 3 3. 2sqrt(3) 4. sqrt(10) 5. 2

If x in[-1/2,1] then sin^(-1)(sqrt(3)/(2)x-1/2sqrt(1-x^(2)))

The orthocentre of the triangle formed by the lines x y=0 and x+y=1 is (a) (1/2,1/2) (b) (1/3,1/3) (c) (0,0) (d) (1/4,1/4)

The circle with centre (3/2,1/2) and radius sqrt(3/2) is

The line y=x is tangent at (0, 0) to a circle of radius 1. The centre of circle may be: (a) (1/(sqrt(2)),-1/(sqrt(2))) (b) (-1/2,1/2) (c) (1/2,-1/2) (d) (-1/(sqrt(2)),1/(sqrt(2)))

The direction ratios of the line 2y+z-5=0=x-3y-6 are proportional to 3,\ 1,\ -2 b. 2,\ -4,\ 1 c. 3/(sqrt(14)),1/(sqrt(14)),(-2)/(sqrt(14)) d. 2/(sqrt(41)),(-4)/(sqrt(41)),1/(sqrt(41))

The orthocentre of the triangle formed by the lines x y=0 and x+y=1 is (1/2,1/2) (b) (1/3,1/3) (0,0) (d) (1/4,1/4)