A circle which passes through origin and cuts intercepts on axes $a$ and $b$. the equation of circle is
A. ${{x}^{2}}+{{y}^{2}}-ax-by=0$
B. $x^{2}+y^{2}+a x+b y=0$
C. ${{x}^{2}}+{{y}^{2}}-ax+by=0$
D. $x^{2}+y^{2}+a x-b y=0$

Hint: The equation for a circle has the generic form: ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$. The coordinates of the circle's center 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 the square formula.

Complete step by step solution:
Given that the passes through origin and cuts intercepts on axes $a$ and $b$.
The radius of a circle is the length of the straight line that connects the center to any point on its circumference. Because a circle's circumference can contain an endless number of points, a circle can have more than one radius. This indicates that a circle has an endless number of radii and that each radius is equally spaced from the circle's center. When the radius's length varies, the circle's size also changes.
$\therefore$ radius $=\sqrt{\left(\dfrac{a^{2}}{2}\right)+\left(\dfrac{b^{2}}{2}\right)}=\left(\dfrac{1}{2}\right) \sqrt{a^{2}+b^{2}}$
The equation of circle
${{(x-(\dfrac{a}{2}))}^{2}}+{{(y-(\dfrac{b}{2}))}^{2}}=((\dfrac{1}{2}\sqrt{{}}{{a}^{2}}+{{b}^{2}}))$
$x^{2}-a x+\left(\dfrac{a^{2}}{4}+y^{2}-b y+\left(\dfrac{b^{2}}{4}\right)\right)$
$=\left(\dfrac{a^{2}}{4}+\left(\dfrac{b^{2}}{4}\right)\right)$
$\therefore x^{2}+y^{2}-a x-b y=0$

Option ‘A’ is correct

Note: The center of a circle is a location inside the circle that is situated in the middle of the circumference.The radius of a circle is the constant distance from the circle's center to any point on the circle.A circle's diameter is defined as the segment of a line that connects two locations on the circle and passes through its center.