The Center of circle `(x+2)^2 + (y+4)^2 = 16` is

The centre of circle (x+2)^2 + (y+2)^2=16 is

Find the centre of circle : (x+2)^(2)+(y+4)^(2) = 16

Find the centre of circle : (x+2)^(2)+(y+4)^(2) = 16

The area of the circle x ^(2) + y ^(2) = 16 "Using Integration ""___________"

Find the area of the circle x ^(2) + y ^(2) = 16

The center of circle x^2+y^2-6x+4y-12=0 is (a,b) then (2a + 3b) is

A circle S passes through the point (0, 1) and is orthogonal to the circles (x -1)^2 + y^2 = 16 and x^2 + y^2 = 1. Then (A) radius of S is 8 (B) radius of S is 7 (C) center of S is (-7,1) (D) center of S is (-8,1)

The condition for a line y = 2x + c to touch the circle x^(2) + y^(2) = 16 is

Find the equation of line joining the center of the circles x^(2)+y^(2)-2x+4y+1=0 and 2x^(2)+2y^(2)-2y+4x+1=0