Radius of circle which is drawn on a normal chord of `y^2 = 4x` as diameter and it pass through vertex is

The radius of the circle, which touches the parabola y^2 = 4x at (1,2) and passes through the origin is:

The locus of mid - points of all chords of parabola y^(2)=4x, for which all cirlces drawn taking them as diameters passes through the vertex of the parabola is a conic whose length of the smallest focal chord is equal to

The locus of mid - points of all chords of parabola y^(2)=4x, for which all cirlces drawn taking them as diameters passes through the vertex of the parabola is a conic whose length of the smallest focal chord is equal to

A circle of radius 4 drawn on a chord of the parabola y^(2)=8x as diameter touches the axis of the parabola. Then the slope of the chord is

A circle of radius 4 drawn on a chord of the parabola y^(2)=8x as diameter touches the axis of the parabola. Then the slope of the chord is

A circle of radius r drawn on a chord of the parabola y^(2)=4ax as diameter touches theaxis of the parabola.Prove that the slope of the chord is (2a)/(r)

Tangents are drawn to the parabola y^2=4x at the point P which is the upper end of latusrectum . Radius of the circle touching the parabola y^2=4x at the point P and passing through its focus is

Tangents are drawn to the parabola y^2=4x at the point P which is the upper end of latusrectum . Radius of the circle touching the parabola y^2=4x at the point P and passing through its focus is

A circle is drawn on the chord of a circle x^(2) + y^(2) =a^(2) as diameter. The chord lies on the line x+y = a . What is the equation of the circle ?