A circle touches both the axes and its centre lies in the fourth quadrant. If its radius is 1 then its equation will be -

Question

A circle touches both the axes and its centre lies in the fourth quadrant. If its radius is 1 then its equation will be - (A) x2+y2-2 x+2 y+1=0 (B) x2+y2+2 x-2 y-1=0 (C) x2+y2-2 x-2 y+1=0 (D) x2+y2+2 x-2 y+1=0

Tardigrade Admin · Accepted Answer

Centre (1,-1) equation (x-1)2+(y+1)2=1

Q. A circle touches both the axes and its centre lies in the fourth quadrant. If its radius is 1 then its equation will be -

$x^{2} + y^{2} - 2 x + 2 y + 1 = 0$

$x^{2} + y^{2} + 2 x - 2 y - 1 = 0$

$x^{2} + y^{2} - 2 x - 2 y + 1 = 0$

$x^{2} + y^{2} + 2 x - 2 y + 1 = 0$

Centre $(1, - 1)$
equation $(x - 1)^{2} + (y + 1)^{2} = 1$

Q. A circle touches both the axes and its centre lies in the fourth quadrant. If its radius is 1 then its equation will be -

x2+y2−2x+2y+1=0

x2+y2+2x−2y−1=0

x2+y2−2x−2y+1=0

x2+y2+2x−2y+1=0

Centre (1,−1) equation (x−1)2+(y+1)2=1

$x^{2} + y^{2} - 2 x + 2 y + 1 = 0$

$x^{2} + y^{2} + 2 x - 2 y - 1 = 0$

$x^{2} + y^{2} - 2 x - 2 y + 1 = 0$

$x^{2} + y^{2} + 2 x - 2 y + 1 = 0$

Centre $(1, - 1)$
equation $(x - 1)^{2} + (y + 1)^{2} = 1$