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Nadusha1986 [10]
3 years ago
9

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

Mathematics
2 answers:
AnnyKZ [126]3 years ago
5 0
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.
Vadim26 [7]3 years ago
4 0

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:

<u><em>Note:</em></u> 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}

<em>First case:</em>

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

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

<em>Second case:</em>

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

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

<em>Third case:</em>

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

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

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