The radical centre of three circles : `x^2 + y^2 + x + 2y + 3 = 0, x^2 + y^2 + 2x + 4y + 5 = 0 and x^2 + y^2 - 7x - 8y - 9 = 0` is : (A) `(- 2/3, - 2/3)` (B) `(1/3, 1/3)` (C) `(1/4, 1/4)` (D) `(0, 0)`

Let circle cuts ortholognally each of the three circles x^(2) + y^(2) + 3x + 4y + 11 = 0, x^(2) + y^(2) - 3x + 7y - 1 = 0 and x^(2) + y^(2) + 2x = 0

Equation of radical axis of the circles x^(2) + y^(2) - 3x - 4y + 5 = 0 and 2x^(2) + 2y^(2) - 10x - 12y + 12 = 0 is

x^(2) - 1 = 0, y^(2) + 4y + 3 = 0

The radical centre of the circles x^(2)+y^(2)=9 , x^(2)+y^(2)-2x-2y-5=0 , x^(2)+y^(2)+4x+6y-19=0 is A) (0,0) (B) (1,1) (C) (2,2) (D) (3,3)

if the radical centre of the three circles x^(2)+y^(2)-2x-1=0 and x^(2)+y^(2)-3y=1 and 2x^(2)+2y^(2)-x-7y-2=0 is Q then Q_(x)+Q_(y) is

(x+2y +1) dx - (2x + 4y +3)dy = 0, x+2y = u

Find the centre and radius of the circle 3x ^(2) + (a + 1) y ^(2) + 6x -9y + a + 4 =0

y-1=m_1(x-3) and y - 3 = m_2(x - 1) are two family of straight lines, at right angled to each other. The locus of their point of intersection is: (A) x^2 + y^2 - 2x - 6y + 10 = 0 (B) x^2 + y^2 - 4x - 4y +6 = 0 (C) x^2 + y^2 - 2x - 6y + 6 = 0 (D) x^2 + y^2 - 4x - by - 6 = 0

The equation of the circle passing through (1/2, -1) and having pair of straight lines x^2 - y^2 + 3x + y + 2 = 0 as its two diameters is : (A) 4x^2 + 4y^2 + 12x - 4y - 15 = 0 (B) 4x^2 + 4y^2 + 15x + 4y - 12 = 0 (C) 4x^2 + 4y^2 - 4x + 8y + 5 = 0 (D) none of these