The intercept on the line `y=x` by the circle `x^2+y^2-2x=0` is AB. Equation of the circle with AB as a diameter is (A) `(x-1/2)^3+(y-1/2)^2=1/2` (B) `(x-1/2)^2+(y-1/2)^2=1/4` (C) `(x+1/2)^2+(y+1/2)^2=1/2` (D) `(x+1/2)^2+(y+1/2)^2=1/4`

The intercept on the line y=x by the circle x^(2)+y^(2)-2x=0 is AB. Show that the equation of the circle with AB as diameter is x^(2)+y^(2)-x-y=0 .

The equation of the tangent to the circle x^(2)+y^(2)-4x+4y-2=0 at (1,1) is

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

If the circle x^2+y^2+2x+3y+1=0 cuts x^2+y^2+4x+3y+2=0 at A and B , then find the equation of the circle on A B as diameter.

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 and b are

The equation of the curve passing through origin, whose slope at any point is (x(1+y))/(1+x^2) , is (A) (1+y)^2-x^2=1 (B) x^2+(y+1)^2=1 (C) (x+y)y=1-x^2 (D) x=ye^((1+y))

A circle of constant radius a passes through the origin O and cuts the axes of coordinates at points P and Q . Then the equation of the locus of the foot of perpendicular from O to P Q is (A) (x^2+y^2)(1/(x^2)+1/(y^2))=4a^2 (B) (x^2+y^2)^2(1/(x^2)+1/(y^2))=a^2 (C) (x^2+y^2)^2(1/(x^2)+1/(y^2))=4a^2 (D) (x^2+y^2)(1/(x^2)+1/(y^2))=a^2

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

The length of the common chord of the circles x^(2)+y^(2)-2x-1=0 and x^(2)+y^(2)+4y-1=0 , is

If tangents are drawn to the ellipse x^2+2y^2=2, then the locus of the midpoint of the intercept made by the tangents between the coordinate axes is (a) 1/(2x^2)+1/(4y^2)=1 (b) 1/(4x^2)+1/(2y^2)=1 (c) (x^2)/2+y^2=1 (d) (x^2)/4+(y^2)/2=1