The equation of the circle passing through the point (1, 0) and (0, 1) and having the smallest radius is -

Question

The equation of the circle passing through the point (1, 0) and (0, 1) and having the smallest radius is - (A) x2 + y2 - 2x - 2y + 1 = 0 (B) x2 + y2 - x - y = 0 (C) x2 + y2 + 2x + 2y - 7 = 0 (D) x2 + y2 + x + y - 2 = 0

Tardigrade Admin · Accepted Answer

Circle whose diametric end points are (1, 0) and (0, 1) will be of smallest radius. (x - 1)(x - 0) + (y - 0) (y - 1) = 0 x2 + y2 - x - y = 0

Q. The equation of the circle passing through the point (1, 0) and (0, 1) and having the smallest radius is -

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

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

$x^{2} + y^{2} + 2 x + 2 y - 7 = 0$

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

Circle whose diametric end points are (1, 0) and (0, 1) will be of smallest radius.
$(x - 1) (x - 0) + (y - 0) (y - 1) = 0$
$x^{2} + y^{2} - x - y = 0$

Q. The equation of the circle passing through the point (1, 0) and (0, 1) and having the smallest radius is -

x2+y2−2x−2y+1=0

x2+y2−x−y=0

x2+y2+2x+2y−7=0

x2+y2+x+y−2=0

Circle whose diametric end points are (1, 0) and (0, 1) will be of smallest radius. (x−1)(x−0)+(y−0)(y−1)=0 x2+y2−x−y=0

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

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

$x^{2} + y^{2} + 2 x + 2 y - 7 = 0$

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

Circle whose diametric end points are (1, 0) and (0, 1) will be of smallest radius.
$(x - 1) (x - 0) + (y - 0) (y - 1) = 0$
$x^{2} + y^{2} - x - y = 0$