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Which point lies on the circle represented by the equation [tex]$x^2 + (y - 12)^2 = 25^2$[/tex]?

A. [tex]$(20, -3)$[/tex]
B. [tex][tex]$(-7, 24)$[/tex][/tex]
C. [tex]$(0, 13)$[/tex]
D. [tex]$(-25, -13)$[/tex]


Sagot :

To determine which point lies on the circle represented by the equation [tex]\( x^2 + (y-12)^2 = 25^2 \)[/tex], we need to substitute each point into the equation and check if it holds true.

The equation of the circle is given by:
[tex]\[ x^2 + (y-12)^2 = 25^2 \][/tex]
which simplifies to:
[tex]\[ x^2 + (y-12)^2 = 625 \][/tex]

Let's verify each point:

Point A: [tex]\((20, -3)\)[/tex]

Substitute [tex]\( x = 20 \)[/tex] and [tex]\( y = -3 \)[/tex]:
[tex]\[ 20^2 + (-3 - 12)^2 = 20^2 + (-15)^2 = 400 + 225 = 625 \][/tex]
The left-hand side equals the right-hand side of the equation [tex]\(625 = 625\)[/tex].

Point B: [tex]\((-7, 24)\)[/tex]

Substitute [tex]\( x = -7 \)[/tex] and [tex]\( y = 24 \)[/tex]:
[tex]\[ (-7)^2 + (24 - 12)^2 = 49 + 12^2 = 49 + 144 = 193 \][/tex]
The left-hand side does not equal the right-hand side [tex]\(193 \neq 625\)[/tex].

Point C: [tex]\((0, 13)\)[/tex]

Substitute [tex]\( x = 0 \)[/tex] and [tex]\( y = 13 \)[/tex]:
[tex]\[ 0^2 + (13 - 12)^2 = 0 + 1 = 1 \][/tex]
The left-hand side does not equal the right-hand side [tex]\(1 \neq 625\)[/tex].

Point D: [tex]\((-25, -13)\)[/tex]

Substitute [tex]\( x = -25 \)[/tex] and [tex]\( y = -13 \)[/tex]:
[tex]\[ (-25)^2 + (-13 - 12)^2 = 625 + (-25)^2 = 625 + 625 = 1250 \][/tex]
The left-hand side does not equal the right-hand side [tex]\(1250 \neq 625\)[/tex].

According to these calculations, only point A (20, -3) lies on the circle.

Therefore, the correct answer is:

A. [tex]\((20, -3)\)[/tex]