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)))`

(1) x^(2)-(sqrt(2)+1)x+sqrt(2)=0

A(1/sqrt(2), 1/sqrt(2)) is a point on the circle x^2 + y^2 = 1 and B is another point on the circle such that AB = pi/2 units. Then coordinates of B can be : (A) (1/sqrt(2), - 1/sqrt(2)) (B) (-1/sqrt(2), 1/sqrt(2)) (C) (-1/sqrt(2), - 1/sqrt(2)) (D) none of these

A((1)/(sqrt(2)),(1)/(sqrt(2))) is a point on the circle x^(2)+y^(2)=1 and B is another point on the circle such that are length AB=(pi)/(2) units. Then,the coordinates of B can be ((1)/(sqrt(2)),1sqrt(2))( b) (-(1)/(sqrt(2)),1sqrt(2))(-(1)/(sqrt(2)),-(1)/(sqrt(2)))( d) none of these

Find the equation of the normal to the circle x^(2)+y^(2)=9 at the point ((1)/(sqrt(2)),(1)/(sqrt(2)))

Tangents drawn from (2, 0) to the circle x^2 + y^2 = 1 touch the circle at A and B . Then (A) A-= (1/2, sqrt(3)/2) , B -= (-1/2, - sqrt(3)/2) (B) A-= (- 1/2, (-sqrt(3)/2) , B -= (1/2, sqrt(3)/2) (C) A-= (1/2, sqrt(3)/2) , B-= (1/2, - sqrt(3)/2) (D) A-= (1/2, - sqrt(3)/2), B-= (1/2, sqrt(3)/2)

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

The image of the centre of the circle x^2 + y^2 = 2a^2 with respect to the line x + y = 1 is : (A) (sqrt(2), sqrt(2) (B) (1/sqrt(2) , sqrt(2) ) (C) (sqrt(2), 1/sqrt(2)) (D) none of these

Let C be the circle with centre (1,1) and radius 1. If T is the circle centered at (0,k) passing through origin and touching the circle C externally then the radius of T is equal to (a) (sqrt(3))/(sqrt(2)) (b) (sqrt(3))/(2) (c) (1)/(2) (d) (1)/(4)

The value of (lim)_(x->0)(sqrt(2)-sqrt(1+cos x))/(sin^2x) is 1/(2sqrt(2)) b. 1/(8sqrt(2)) c. 1/(4sqrt(2))\ d. -1/(4sqrt(2))