The equation of the circle in the first quadrant touching each coordinate axis at a distance of one unit from the origin is

Question

The equation of the circle in the first quadrant touching each coordinate axis at a distance of one unit from the origin is (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=0 (D) x2+y2-2 x+2 y-1=0

Tardigrade Admin · Accepted Answer

(Since, the circle is in the first quadrant touching each co-ordinate axis at a distance of one unit from origin.

∴

It passes through the points

(1, 0)

and

(0, 1)

.
Also, its centre will be

(1, 1)

and radius is one unit. Hence, the equation of required circle is

(x - 1)^{2} + (y - 1)^{2} = 1^{2}

Q. The equation of the circle in the first quadrant touching each coordinate axis at a distance of one unit from the origin is

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

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

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

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

Q. The equation of the circle in the first quadrant touching each coordinate axis at a distance of one unit from the origin is

x2+y2−2x−2y+1=0

x2+y2−2x−2y−1=0

x2+y2−2x−2y=0

x2+y2−2x+2y−1=0

(Since, the circle is in the first quadrant touching each co-ordinate axis at a distance of one unit from origin. ∴ It passes through the points (1,0) and (0,1). Also, its centre will be (1,1) and radius is one unit. Hence, the equation of required circle is (x−1)2+(y−1)2=12

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

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

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

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