The equation of the circle which touches both the axis and whose centre is $({{x}_{1}},{{y}_{1}})$ is
A) ${{x}^{2}}+{{y}^{2}}+2{{x}_{1}}(x+y)+x_{1}^{2}=0$
B) $x^{2}+y^{2}-2 x_{1}(x+y)+x_{1}^{2}=0$
C) ${{x}^{2}}+{{y}^{2}}=x_{1}^{2}+y_{1}^{2}$
D) $\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{xx}_{1}+2 y \mathrm{y}_{1}=0$

Hint: We have given the centre point of the circle to find out the equation of the circle. We know the general equation of a circle and we have given the centre points. So by substituting the values of points we can easily find the equation of the circle.

Complete step by step Solution:
Let the radius be $y$
Then,
Equation is
${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$
The location of a circle in the Cartesian plane is represented by a circle equation. We can write the equation for a circle if we know the location of the circle's centre and how long its radius is. All of the points on the circle's circumference are represented by the circle equation.
The group of points whose separation from a fixed point has a constant value is represented by a circle. The radius of the circle abbreviated $r$, is a constant that describes this fixed point, which is known as the circle's centre. The formula for a circle$\left(\mathrm{x}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}-\mathrm{y}_{1}\right)^{2}=\mathrm{r}^{2}$ whose centre is at $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$

$\therefore\left(x-x_{1}\right)^{2}+\left(y-y_{1}\right)^{2}=y_{1}^{2}$
$x^{2}+x_{1}^{2}-2 x_{1}+y^{2}+y_{1}^{2}-2 y_{1}=y_{1}^{2}$
$\therefore x^{2}+y^{2}-2 x_{1}-2 y_{1}+x_{1}^{2}=0$
$\left[\therefore \mathrm{x}_{1}=\mathrm{y} 1\right]$
$\therefore x^{2}+y^{2}-2 x_{1}-2 y x_{1}+x_{1}^{2}=0$
$\therefore x^{2}+y^{2}-2 x_{1}(x+y)+x_{1}^{2}=0$

Therefore, the correct option is (B).


Note: Note that the equation for a circle has the generic form: ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$. The coordinates of the circle's centre and radius are found using this general form, where g, f, and c are constants. The general form of the equation of a circle makes it difficult to identify any significant properties about any specific circle, in contrast to the standard form, which is simpler to comprehend. So, to quickly change from the generic form to the standard form, we will use the completing square formula.