Question

Equation of circle inscribed in $|x-a|+|y-b|=1$ is

• A. $(x+a)^{2}+(y+b)^{2}=2$
• B. $(x-a)^{2}+(y-b)^{2}=\frac{1}{2}$
• C. $(x-a)^{2}+(y-b)^{2}=\frac{1}{\sqrt{2}}$
• D. $(x-a)^{2}+(y-b)^{2}=1$

Answer 1

Answer: (B)

$|x-a|+|y-b|=1$ is square with center at $(a, b)$ which is center of circle also.
Distance between two parallel sides of square is $\sqrt{2}$.
So, radius of the required circle is $\frac{1}{\sqrt{2}}$.
Hence equation is $(x-a)^{2}+(y-b)^{2}=\frac{1}{2}$