Question

Does the point (2 3 , 2) lie on the circle that is centered at the origin and contains the point (0, -4)? Why?

Answer 1

Consider this option:
1.the equation of the circle using its centre and point (0;-4) is:
x²+y²=4.
2. to substitute given point (23;2) into the equation of the circle:
23²+2²≠4

answer: this point does not belongs to the circle.

Answer 2

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

$C=(a,b)$

and its radius is $r$, the ordinary equation of this circle is given by:

$(x-a)^2+(y-b)^2=r^2$

Since the circle is centered at the origin:

$C=(a,b)=(0,0)\\\\Hence\\\\(x-0)^2+(y-0)^2=r^2\\\\x^2+y^2=r^2$

Now, let's find $r$ using the data provided. Evaluating the point (0,-4) into the equation:

$(0)^2+(-4)^2=r^2\\\\16=r^2\\\\r=\pm 4$

Thus the equation for the circle given by the problem is:

$x^2+y^2=16$

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:

$2\hspace{3} 3=23\\2\hspace{3} 3=2*3=6\\2\hspace{3} 3=\frac{2}{3}$

First case:

$(23)^2+(2)^2=16\\\\533\neq16$

It doesn't satisfy the equation, therefore doesn't lie on the circle.

Second case:

$(6)^2+(2)^2=16\\\\40\neq16$

It doesn't satisfy the equation, therefore doesn't lie on the circle.

Third case:

$(\frac{2}{3} )^2+(2)^2=16\\\\\frac{40}{9} \neq16$

It doesn't satisfy the equation, therefore doesn't lie on the circle.