If the line `a x+b y=2` is a normal to the circle `x^2+y^2-4x-4y=0` and a tangent to the circle `x^2+y^2=1` , then `a=1/2,b=1/2` `a=(1+sqrt(7))/2` , `b=(1+sqrt(7))/2` `a=1/4,b=3/4` (d) `a=1,b=sqrt(3)`

If the line ax+by=2 is a normal to the circle x^(2)+y^(2)-4x-4y=0 and a tangent to the circle x^(2)+y^(2)=1, then a and bare

Find the equation of the tangent and normal to the circle 2x^2+2y^2-3x-4y+1=0 at the point (1,2)

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

If the straight line ax+by=2;a,b!=0 touches the circle x^(2)+y^(2)-2x=3 and is normal to the circle x^(2)+y^(2)-4y=6 ,then the values of 'a' and 'b'are?

The tangents to the circle x^(2)+y^(2)-4x+2y+k=0 at (1,1) is x-2y+1=0 then k=

Prove that the equation of any tangent to the circle x^(2)+y^(2)-2x+4y-4=0 is of the form y=m(x-1)+3sqrt(1+m^(2))-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)

Area of the triangle formd by the line x+y=3 and the angle bisector of the line pair x^(2)-y^(2)+4y-4=0 is : a.1/2 b.1 c.3/2d

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 3sqrt(5)(b)5sqrt(3)(c)2sqrt(5)(d)5sqrt(2)