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)))`

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 A B=pi/2 units. Then, the 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

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

The line 2x-y+1=0 is tangent to the circle at the point (2, 5) and the center of the circle lies on x-2y=4. The radius of the circle is (a) 3sqrt(5) (b) 5sqrt(3) (c) 2sqrt(5) (d) 5sqrt(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 incenter of the triangle with vertices (1,sqrt(3)),(0,0), and (2,0) is (a) (1,(sqrt(3))/2) (b) (2/3,1/(sqrt(3))) (c) (2/3,(sqrt(3))/2) (d) (1,1/(sqrt(3)))

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

A B C D is a square of unit area. A circle is tangent to two sides of A B C D and passes through exactly one of its vertices. The radius of the circle is (a) 2-sqrt(2) (b) sqrt(2)-1 (c) sqrt(2)-1/2 (d) 1/(sqrt(2))

sin^(-1)(2xsqrt(1-x^2))=2sin^(-1)x is true if x in [0,\ 1]\ b. [-1/(sqrt(2)),1/(sqrt(2))] c. [-1/2,1/2] d. [-(sqrt(3))/2,(sqrt(3))/2]

If y = mx + c is a tangent to the circle (x – 3)^2 + y^2 = 1 and also the perpendicular to the tangent to the circle x^2 + y^2 = 1 at (1/sqrt(2),1/sqrt(2)) , then

A circle S of radius ' a ' is the director circle of another circle S_1,S_1 is the director circle of circle S_2 and so on. If the sum of the radii of all these circle is 2, then the value of ' a ' is (a) 2+sqrt(2) (b) 2-1/(sqrt(2)) (c) 2-sqrt(2) (d) 2+1/(sqrt(2))