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` . Then find the radius of the circle.

The line 2x-y+1=0 is tangent to the circle at the point (2, 5) and the centre of the circle lies on x-2y=4. The square of the radius of the circle is

The line 2x-y+1=0 is tangent to the circle at the point (2,5) and the centre of the circles lies on x-2y=4. The radius of the circle is

The line 2x-y+1=0 is tangent to the circle at the point (2,5) and the centre of the circles lies on x-2y=4. The radius of the circle is :

The line 2x-y+1=0 is tangent to the circle at the point (2,5) and the centre of the circles lies on x-2y=4 The radius of the circle is:

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)

The line 2x-y+1=0 touches a circle at the point (2, 5) and the centre of the circle lies on x-2y=4 . The diameter (in units) of the circle is

If the line 2x-y+1=0 touches the circle at the point (2,5) and the centre of the circle lies in the line x+y-9=0. Find the equation of the circle.

The line 5x-y=3 is a tangent to a circle at the point (2,7) and its centre is on theline x+2y=19. Find the equation of the circle.

A circle touches the lines x-y- 1 =0 and x -y +1 =0. the centre of the circle lies on the line

If the line y=2-x is tangent to the circle S at the point P(1,1) and circle S is orthogonal to the circle x^(2)+y^(2)+2x+2y-2=0 then find the length of tangent drawn from the point (2,2) to the circle S