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

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

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

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

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

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Solution

The correct option is C $x^{2} + y^{2} - x - y = 0$
The radius will be smallest if the given points are diametric end points.

$\Rightarrow d i a m e t e r = d = \sqrt{1^{2} + 1^{2}}$

$= \sqrt{2} .$ $r = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2} .}$

$C e n t r e \Rightarrow (\frac{1}{2}, \frac{1}{2})$ the midpoint of the given points

$\Rightarrow (x - \frac{1}{2})^{2} + (y \frac{- 1}{2})^{2} = \frac{1}{2}$

$\Rightarrow x^{2} + y^{2} + \frac{1}{4} + \frac{1}{4} - x - y = \frac{1}{2} .$

$\Rightarrow x^{2} + y^{2} - x - y = 0.$ $A n s .$

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

The correct option is C x2+y2−x−y=0The radius will be smallest if the given points are diametric end points.⇒diameter=d=√12+12 =√2. r=√22=1√2. Centre⇒(12,12) the midpoint of the given points ⇒(x−12)2+(y−12)2=12 ⇒x2+y2+14+14−x−y=12. ⇒x2+y2−x−y=0. Ans.

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