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

No, because it doesn't satisfy the equation of the circumference

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=(2)/(3)`

First case:

`(23)^2+(2)^2=16\\\\533\\eq16`

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

Second case:

`(6)^2+(2)^2=16\\\\40\\eq16`

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

Third case:

`((2)/(3) )^2+(2)^2=16\\\\(40)/(9) \\eq16`

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