`y^(2)=4x` और `x^(2)=4y`. प्रतिच्छेदन कोण

y^(2)=4x, y^(2)=4(4-x)

(3x - 4y)^(2) - (4x + 3y)^(2) = "______" .

Statement 1: The equations of the straight lines joining the origin to the points of intersection of x^(2)+y^(2)-4x-2y=4 and x^(2)+y^(2)-2x-4y-4=0 is x-y=0 . Statement 2: y+x=0 is the common chord of x^(2)+y^(2)-4x-2y=4 and x^(2)+y^(2)-2x-4y-4=0

The equation of the circle passing through the point of intersection of the circles x^(2)+y^(2)-4x-2y=8 and x^(2)+y^(2)-2x-4y=8 and (-1,4) is (a)x^(2)+y^(2)+4x+4y-8=0(b)x^(2)+y^(2)-3x+4y+8=0(c)x^(2)+y^(2)+x+y=0(d)x^(2)+y^(2)-3x-3y-8=0

The centre of circle cutting the three circles x^(2)+y^(2)-4x-4y+5=0 , x^(2)+y^(2)-4x-6y+5=0 , x^(2)+y^(2)-4x-8y+5=0 orthogonally is

If r_(1), r_(2) " and " r_(3) are the radii of the circle x^(2) + y^(2) - 4x + 6y = 5 , x^(2) + y^(2) + 6x - 4y = 3 " and " x^(2) + y^(2) - 2x + 4y = 8

Prove that the centres of the circle x^(2) + y^(2) - 4x - 2y + 4 = 0, x^(2) + y^(2) - 2x - 4y + 1 = 0 and x^(2) + y^(2) + 2x - 8y + 1 = 0 are collinear. More over prove that their radii are in geometric pregression.