If the radical center of the given three circles x2 + y2 = 1, x2 + y2 -2x - 3 =0 and x2 + y2 -2y - 3 = 0 is C(α,β) and r is the sum of the radii of the given circles, then the circle with C(α,β) as center and r as radius is
(x - 1)2 + (y - 1)2 = 2
(x - 1)2 + (y + 1)2 =4
(x - 2)2 + (y - 2)2 = 25
(x + 1)2 + (y + 1)2 = 25
The correct option is: (D) (x + 1)2 + (y + 1)2 = 25
Let \( F \) and \( F' \) be the foci of the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) (where \( b<2 \)), and let \( B \) be one end of the minor axis. If the area of the triangle \( FBF' \) is \( \sqrt{3} \) sq. units, then the eccentricity of the ellipse is:
A common tangent to the circle \( x^2 + y^2 = 9 \) and the parabola \( y^2 = 8x \) is
If the equation of the circle passing through the points of intersection of the circles \[ x^2 - 2x + y^2 - 4y - 4 = 0, \quad x^2 + y^2 + 4y - 4 = 0 \] and the point \( (3,3) \) is given by \[ x^2 + y^2 + \alpha x + \beta y + \gamma = 0, \] then \( 3(\alpha + \beta + \gamma) \) is: