Question

Equation of circle touching all the sides of the square $|x-2|+|y-3|=4$ will be

• A. $(x-2{)}^2+(y-3{)}^2=2\sqrt{2}$
• B. $(x-2{)}^2+(y-3{)}^2=4$
• C. $(x-2{)}^2+(y-2{)}^2=10$
• D. $(x-2{)}^2+(y-3{)}^2=8$

Answer 1

The correct option is D $(x-2{)}^2+(y-3{)}^2=8$

equation of circle = (?)

$|x-2|+|y-3|=4$

$\to |x-4|+|y-4|=1$ will represent square in the xy plare

centered at (a,b) of side length being $\sqrt{2}$ units.

Radius of circle $=\frac{1}{2}\times$ side length of square $=\frac{1}{2}\times \sqrt{2}=1units.$

$=\frac{1}{\sqrt{2}}\times \left(4\right)$

$=2\sqrt{2}$

$\to$ Hence equation of circle

$(x-a{)}^2+(y-b{)}^2=r^2$

$(x-2{)}^2+(y-3{)}^2=(2\sqrt{2}{)}^2$

$(x-2{)}^2+(y-3{)}^2=8$