Answer:
No, because it doesn't satisfy the equation of the circumference
Step-by-step explanation:
A circle is the locus of points on the plane that are equidistant from a fixed point called the center. For a circle whose center is the point
[tex]C=(a,b)[/tex]
and its radius is [tex]r[/tex], the ordinary equation of this circle is given by:
[tex](x-a)^2+(y-b)^2=r^2[/tex]
Since the circle is centered at the origin:
[tex]C=(a,b)=(0,0)\\\\Hence\\\\(x-0)^2+(y-0)^2=r^2\\\\x^2+y^2=r^2[/tex]
Now, let's find [tex]r[/tex] using the data provided. Evaluating the point (0,-4) into the equation:
[tex](0)^2+(-4)^2=r^2\\\\16=r^2\\\\r=\pm 4[/tex]
Thus the equation for the circle given by the problem is:
[tex]x^2+y^2=16[/tex]
In order to corroborate if the the point (2 3, 2) lie on the circle, we need to evaluate it into the equation and check if it satisfy the equation:
Note: I don't know what you mean with 2 3, so I will assume 3 cases:
[tex]2\hspace{3} 3=23\\2\hspace{3} 3=2*3=6\\2\hspace{3} 3=\frac{2}{3}[/tex]
First case:
[tex](23)^2+(2)^2=16\\\\533\neq16[/tex]
It doesn't satisfy the equation, therefore doesn't lie on the circle.
Second case:
[tex](6)^2+(2)^2=16\\\\40\neq16[/tex]
It doesn't satisfy the equation, therefore doesn't lie on the circle.
Third case:
[tex](\frac{2}{3} )^2+(2)^2=16\\\\\frac{40}{9} \neq16[/tex]
It doesn't satisfy the equation, therefore doesn't lie on the circle.
