A tangent PT is drawn to the circle `x^2+y^2=4` at point `p(sqrt3, 1)` A straight line L perpendicular to PT is a tangent to the circle `(x-3)^2 + y^2 =1` A possible equation of `L` is

A tangent PT is drawn to the circle x^2 + y^2= 4 at the point P(sqrt3,1). A straight line L is perpendicular to PT is a tangent to the circle (x-3)^2 + y^2 = 1 Common tangent of two circle is: (A) x=4 (B) y=2 (C) x+(sqrt3)y=4 (D) x+2(sqrt2)y=6

A tangent PT is drawn to the circle x^(2)+y^(2)=4 at the point P(sqrt(3),1). A straight line L, perpendicular to PT is a tangent to the circle (x-3)^(2)+y^(2)=1 then find a common tangent of the two circles

If a line, y = mx + c is a tangent to the circle, (x-1)^2 + y^2 =1 and it is perpendicular to a line L_1 , where L_1 is the tangent to the circle x^2 + y^2 = 8 at the point (2, 2), then :

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

show that the tangent of the circle x^(2)+y^(2)=25 at the point (3,4) and (4,-3) are perpendicular to each other.

Locus of point of intersection of perpendicular tangents to the circle x^(2)+y^(2)-4x-6y-1=0 is

If the tangent from a point p to the circle x^(2)+y^(2)=1 is perpendicular to the tangent from p to the circle x^(2)+y^(2)=3, then the locus of p is