Question

The equation of the circle which passes through $(1,0)$ and $(0,1)$ and has its radius as small as possible is

• A. $x^2+y^2+x+y=0$
• B. $x^2+y^2-x+y=0$
• C. $x^2+y^2+x-y=0$
• D. $x^2+y^2-x-y=0$

Answer 1

The correct option is D $x^2+y^2-x-y=0$

Solution

(1,0) and (0,1) must be diametric end points

for minimum possible radius.

$r=\frac{1}{2}\sqrt{1^2+1^2}=\frac{1}{\sqrt{2}}$

centre $\to (\frac{1}{2},\frac{1}{2})$

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

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

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

D is correct.