Question

Find the equation of the circle which touches both axes in the first quadrant and whose radius is a
A )$\quad x^{2}+y^{2}-2 a x-2 a y+a^{2}=0$
B )$\quad x^{2}+y^{2}-2 a x+2 a y-a^{2}=0$
C )$\quad x^{2}+y^{2}+2 a x-2 a y+a^{2}=0$
D ) None of the above

Answer 1

Hint:We have the general equation of circle. And it is given that the circle touches both axes in the first quadrant. So we can find the general equation of the given circle by substituting the a value. The radius of a circle is one of its most important features It is the separation between a point on a circle's edge and its center.

Complete step by step Solution:
Given,
In the first quadrant, a circle with radius a contacts both axes, hence its center will be $(a, a)$.

Therefore the required equation is
$\Rightarrow(x-a)^{2}+(y-a)^{2}=a^{2}$
$\Rightarrow x^{2}+a^{2}-2 a x+y^{2}+a^{2}-2 a y=a^{2}$
$\Rightarrow x^{2}+y^{2}-2 a x-2 a y+2 a^{2}-a^{2}=0$
$\Rightarrow {{x}^{2}}+{{y}^{2}}-2ax-2ay+{{a}^{2}}=0$

Therefore, the correct option is A.

Additional information:
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.

Note: 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 square formula.