Question

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

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

Answer 1

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

Let the equation of the required circle be
$x^2+y^2+2gx+2fy+c=0\dots \left(i\right)$
This passes through $A(1,0)$ and $B(0,1)$.
Therefore,
$1+2g+c=0$ and $1+2f+c=0$
$\Rightarrow g=-\left(\frac{c+1}{2}\right)$ and $f=-\left(\frac{c+1}{2}\right)$
Let $r$ be the radius of circle $\left(i\right)$
Then, radius
$r=\sqrt{g^2+f^2-c}$
$\Rightarrow r=\sqrt{{\left(\frac{c+1}{2}\right)}^2+{\left(\frac{c+1}{2}\right)}^2-c}$
$\Rightarrow r=\sqrt{\frac{c^2+1}{2}}$
$\Rightarrow r^2=\frac{1}{2}(c^2+1)$
Clearly, $r$ is minimum when $c=0$ and the minimum value of $r$ is $\frac{1}{\sqrt{2}}$
For $c=0$, we have
$g=-\frac{1}{2}$ and $f=-\frac{1}{2}$
On substituting the values of $g,f$ and $c$ in equation $\left(i\right)$, we get
$x^2+y^2-x-y=0$