The equation $(x^2-4)^2 + (y^2-9)^2 = 0$ represents points:

Question

The equation $(x^2-4)^2 + (y^2-9)^2 = 0$ represents points:

$a$. Which lie on the circle with centre at $(0,0)$.

$b$. which lie on a circle with centre $(2,3)$

$c$. Which are collinear

$d$. None of the above.

My Attempt:

The book's answer option $a$ but I don't understand how.

Answer 1

Hint:-

$$a^2\ge0\qquad(\text{where $a\in \Bbb{R}$})$$

$$$$

And what case of this inequality do you have?

Answer 2

Square of any real quantity is $\geq 0$. So if sum of two squares is $=0$, it means the two squares themselves have to be zero, because if one of them is non-zero, the sum would be greater than zero. Therefore $x^2=4$ and $y^2=9$ which means the points are $(\pm 2,\pm3)$, which lie on a circle centered at $0$ (the circle $x^2+y^2=13$).