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Which of these points is on the circle [tex]\((x+2)^2 + (y-1)^2 = 25\)[/tex]?

Sagot :

To determine which points lie on the circle defined by the equation [tex]\((x + 2)^2 + (y - 1)^2 = 25\)[/tex], we need to check each given point and see if it satisfies this equation.

The center of the circle is [tex]\((-2, 1)\)[/tex] and the radius squared is 25. So, for a point [tex]\((x, y)\)[/tex] to be on the circle, it must satisfy:

[tex]\[(x + 2)^2 + (y - 1)^2 = 25\][/tex]

Let's check each point:

1. Point (3, 3):
[tex]\[ (3 + 2)^2 + (3 - 1)^2 = 5^2 + 2^2 = 25 + 4 = 29 \][/tex]
Since [tex]\( 29 \neq 25 \)[/tex], the point (3, 3) is not on the circle.

2. Point (3, -3):
[tex]\[ (3 + 2)^2 + (-3 - 1)^2 = 5^2 + (-4)^2 = 25 + 16 = 41 \][/tex]
Since [tex]\( 41 \neq 25 \)[/tex], the point (3, -3) is not on the circle.

3. Point (2, 5):
[tex]\[ (2 + 2)^2 + (5 - 1)^2 = 4^2 + 4^2 = 16 + 16 = 32 \][/tex]
Since [tex]\( 32 \neq 25 \)[/tex], the point (2, 5) is not on the circle.

4. Point (0, -4):
[tex]\[ (0 + 2)^2 + (-4 - 1)^2 = 2^2 + (-5)^2 = 4 + 25 = 29 \][/tex]
Since [tex]\( 29 \neq 25 \)[/tex], the point (0, -4) is not on the circle.

After checking all the points, we find that none of the given points lie on the circle [tex]\((x + 2)^2 + (y - 1)^2 = 25\)[/tex].