For the circle ${{x}^{2}}+{{y}^{2}}+3x+3y=0$, which of the following relations is true?
A. centre lies on X-axis
B. centre lies on Y-axis
C. centre is at origin
D. circle passes through origin
VerifiedHint: We first try to form the given circle in its general form of ${{\left( x-\alpha \right)}^{2}}+{{\left( y-\beta \right)}^{2}}={{r}^{2}}$ to find the centre and the radius. We equate with the general formula of ${{\left( x-\alpha \right)}^{2}}+{{\left( y-\beta \right)}^{2}}={{r}^{2}}$ and find the value of the centre $O\equiv \left( \alpha ,\beta \right)$. We then find the position of the centre with respect to the axes. Also, we place the value of $x=y=0$ to find if it goes through the origin.
Complete step by step answer:
It’s given that the equation of the circle is ${{x}^{2}}+{{y}^{2}}+3x+3y=0$. We transform it in its general form of ${{\left( x-\alpha \right)}^{2}}+{{\left( y-\beta \right)}^{2}}={{r}^{2}}$ and get ${{\left( x+\dfrac{3}{2} \right)}^{2}}+{{\left( y+\dfrac{3}{2} \right)}^{2}}=\dfrac{9}{2}$. O is the centre.
Equating with the general equation of circle ${{\left( x-\alpha \right)}^{2}}+{{\left( y-\beta \right)}^{2}}={{r}^{2}}$, we get the centre as $O\equiv \left( -\dfrac{3}{2},-\dfrac{3}{2} \right)$ and the radius as $\dfrac{3}{\sqrt{2}}$ units.
We can see that the centre doesn’t lies on X-axis nor Y-axis. It is also not the origin.
Last, we check if it passes through the origin.
We put the value of $x=y=0$.
We get ${{x}^{2}}+{{y}^{2}}+3x+3y={{0}^{2}}+{{0}^{2}}+3\times 0+3\times 0=0$. The point $\left( 0,0 \right)$ satisfies the circle which means the circle passes through the origin.
The correct option is D.
Note:
The formula to check if the circle passes through the origin is to check the constant c in the general equation of circle ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$. If the value of c is 0 then we can say that the circle passes through the origin.
