The circle having `2x +y-5=0 and 2x+y+15=0` as tangents and `(-5,-5)` is one of the point of contact of one of them, then the equation of circle is

Find the point of contact of the tangent x+y-7=0 to the circle x^2+y^2-4x-6y+11=0

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

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 :

Show that the line x - 7y + 5 = 0 is a tangent to the circle x^2+y^2-5x+5y=0 . Find the point of contact. Find also the equation of tangent parallel to the given line.

If 5x-12y+10=0 and 12y-5x+16=0 are tangents of a circle, then radius of that circle is

If the circles x ^(2) + y ^(2) + 5x -6y-1=0 and x ^(2) + y^(2) +ax -y +1=0 intersect orthogonally (the tangents at the point of intersection of the circles are at right angles), the value of a is