Show that the curves `x^(2)+y^(2)=2, 3x^(2)+y^(2)=4x` have a common tangent at the point (1,1)

If (-1/3-1) is a centre of similitude for the circles x^(2)+y^(2)=1 and x^(2)+y^(2)_2x-6y-6=0 then the length of common tangent of the circles is

Ilf the circle x^(2)+y^(2)=2 and x^(2)+y^(2)_4x-4y+lamda=0 have exactly three real common tangents then lamda=

The two curves y=x^2+1,y=3x^2-4x+3 at (1,2)

Tangents are drawn from each point on the line 2x+y=4 to the circle x^(2)+y^(2)=4 . Show that the chord of contact pass through a point (1/2,1/4) .

If the length of the tangent from (1,2) to the circle x^(2)+y^2+x+y-4=0 and 3x^(2)+3y^(2)-x-y-lambda=0 are in the ratio 4:3 then lambda=

The equation of the common tangent at the point contact of the circles x^(2) + y^(2) - 10x + 2y + 10 = 0 , x^(2) + y^(2) - 4x - 6y + 12 = 0 is

Find points on the curve (x^(2))/(4) +(y^(2))/(25) =1 at which the tangents are (i) parallel to x-axis (ii) parallel to y-axis.