Solving the quadratic equation $a X^2 + a Y^2 + 2 b X + 2 c Y + d = 0$.

Question

I'm trying to solve the following quadratic equation, but I have no idea how: $$a X^2 + a Y^2 + 2 b X + 2 c Y + d = 0,$$ where the coefficients are real numbers. Can someone give me a hint?

Answer 1

Find solutions of $ax^2+ay^2+2bx+2cy+d=0$, where $a,b,c,d\in\mathbb R$.

Meant to be a comment

Assuming $b^2+c^2\ge d, a\neq 0$, you can always choose $a>0$ and then

\begin{align*} \left(x+\frac ba\right)^2+\left(y+\frac ca\right)^2&=r^2\\ x^2+y^2+\frac {2b}{a}x+\frac {2c}{a}y+\frac {b^2+c^2}{a^2}&=r^2\\ ax^2+ay^2+2bx+2cy+\underbrace{(b^2+c^2-ar^2)}_{d}&=0\\ \end{align*} is a real circle with center $\left(-\frac ba,-\frac ca\right)$ and $r=\frac{\sqrt{b^2+c^2-d}}a$.