Show that the circle on the chord `xcosalpha+ ysinalpha-p = 0` of the circle `x^2+ y^2 = a^2` as diameter is `x^2 + y^2 - a^2 - 2p (xcosalpha + ysinalpha-p) = 0.`

Show that the circle on the chord x cos alpha+y sin alpha-p=0 of the circle x^(2)+y^(2)=a^(2) as diameter is x^(2)+y^(2)-a^(2)-2p(x cos alpha+y sin alpha-p)=0

Equation of the circle on the common chord of the circles x ^(2) + y ^(2) -ax =0 and x ^(2) + y ^(2) - ay =0 as a diameter is

The midpoint of the chord y=x-1 of the circle x^(2)+y^(2)-2x+2y-2=0 is

The equation of the circle described on the common chord of the circles x^(2)+y^(2)+2x=0 and x^(2)+y^(2)+2y=0 as diameter, is

The equation of the circle on the common chord of the circles (x-a)^(2)+y^(2)=a^(2) and x^(2)+(y+b)^(2)=b^(2) as diameter, is

Determine equation of the circle whose diameter is the chord x+y=1 of the circle x^(2)+y^(2)=4

If the straight line xcosalpha+ysinalpha=p touches the curve (x^2)/(a^2)+(y^2)/(b^2)=1 , then prove that a^2\ cos^2alpha+b^2\ s in^2alpha=p^2 .

Equation of any circle passing through the point(s) of intersection of circle S=0 and line L=0 is S + kL = 0 . Let P(x_1, y_1) be a point outside the circle x^2 + y^2 = a^2 and PA and PB be two tangents drawn to this circle from P touching the circle at A and B . On the basis of the above information : The circle which has for its diameter the chord cut off on the line px+qy - 1 = 0 by the circle x^2 + y^2 = a^2 has centre (A) (p/(p^2 + q^2), (-q)/(p^2 + q^2) (B) (p/(p^2 + q^2), (q)/(p^2 + q^2) (C) (p/(p^2 + q^2), (q)/(p^2 + q^2) (D) none of these