Circle through Co normal points

A circle x^(2)+y^(2)+2gx+2fy+C=0 passes through three co normal points on the parabola y^(2)=4ax, also the point of concurrency of the normal is (alpha,beta). Then correct options are?

Statement :1 If a parabola y ^(2) = 4ax intersects a circle in three co-normal points then the circle also passes through the vertr of the parabola. Because Statement : 2 If the parabola intersects circle in four points t _(1), t_(2), t_(3) and t_(4) then t _(1) + t_(2) + t_(3) +t_(4) =0 and for co-normal points t _(1), t_(2) , t_(3) we have t_(1)+t_(2) +t_(3)=0.

Show that the circle through three points the normals at which to the parabola y^(2)=4ax are concurrent at the point (h,k) is 2(x^(2)+y^(2))-2(h+2a)x-ky=0

A normal is drawn to the parabola y^(2)=9x at the point P(4, 6). A circle is described on SP as diameter, where S is the focus. The length of the intercept made by the circle on the normal at point P is :

If a circle passes through four points, then the four points are said to be……………………

Equation of a circle passing through 3points

In parabola y^2=4x, From the point (15,12), three normals are drawn to the three co-normals point is

The radius of the circle passing through the point (6,2) and having x+y=6 as its normal and x+2y=4 as its diameter is :