Answer:
Yes
Explanation:
Both arguments are squared so the equation are in the form of a circle.
A lotus is a point that is equidistant from some fixed point of a conic section. This is a circle so the fixed point is called a center of the circle.
The center of the circle is equidistant from any point on the circumference of the circle. In order for a point to be on the circle, it must satisfy the circle equation.
![{x}^(2) + {y}^(2) = {r}^(2)](https://img.qammunity.org/2022/formulas/mathematics/high-school/gz5fhgdz8x6agyemmiagl690h13xcgsd47.png)
Let plug in what we know
![3 {}^(2) + ( - 4) {}^(2) = {5}^(2)](https://img.qammunity.org/2022/formulas/mathematics/high-school/xfnyjw6gpd9jddg9ppt91hklsso3o03xyf.png)
![9 + 16 = {5}^(2)](https://img.qammunity.org/2022/formulas/mathematics/high-school/n9kq6xu7bdifjprky52tagc7x0xht4ohsn.png)
So 3,-4 is on the locus of the circle.