Question

Find the equation of the circle concentric with $x^{2}+y^{2}-4 x-6 y-3=0$ and which touches the $y$-axis.
A) ${{x}^{2}}+{{y}^{2}}-4x-6y+9=0$
B) $x^{2}+y^{2}-4 x-6 y+18=0$
C) ${{x}^{2}}+{{y}^{2}}-4x-6y-4=0$
D) $x^{2}+y^{2}-4 x-6 y+3=0$

Answer 1

Hint: A circle's centre and radius make up its 2D shape. If we are aware of the circle's center and radius, we can draw any circle. The radii of a circle are infinitely variable. The midpoint where all of the radii meet is the center of a circle. The center of the circle's diameter is another way to describe it.

Complete step by step Solution:
Center of given circle $=\left(\dfrac{4}{2}, \dfrac{6}{2}\right)=(2,3)$
The general equation of a circle is another name for the center of a circle formula. If the radius is r, the center's coordinates are $(h,k),$and any point on the circle is$(x, y)$, the center of the circle formula is as follows:
$(x-h)^{2}+(y-k)^{2}=r^{2}$
The equation for the center of the circle is another name for this. In the sections that follow, we'll use this formula to determine a circle's equation or center
Let $(x-2)^{2}+(y-3)^{2}=r^{2}$ be a concentric circle
Since $x=0$ is a tangent
Radius is the Perpendicular distance from center to tangent
$\Rightarrow \mathrm{r}=\dfrac{|2+0|}{\sqrt{1+0}} \Rightarrow \mathrm{r}=2$
Equation of circle is $(x-2)^{2}+(y-3)^{2}=2^{2}$
$\Rightarrow x^{2}+y^{2}-4 x-6 y+9=0$

Hence, the correct option is (A).

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.