Respuesta :
Answer:
Yes
Step-by-step 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.
[tex] {x}^{2} + {y}^{2} = {r}^{2} [/tex]
Let plug in what we know
[tex]3 {}^{2} + ( - 4) {}^{2} = {5}^{2} [/tex]
[tex]9 + 16 = {5}^{2} [/tex]
So 3,-4 is on the locus of the circle.
