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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 check each point given in the options to see if it satisfies this equation.

The circle equation is:
[tex]\[x^2 + (y - 12)^2 = 625\][/tex]
since [tex]\(625\)[/tex] is [tex]\(25^2\)[/tex].

Let's check each point:

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

Substitute [tex]\(x = 20\)[/tex] and [tex]\(y = -3\)[/tex] into the equation:
[tex]\[20^2 + (-3 - 12)^2 = 400 + (-15)^2 \][/tex]
[tex]\[ = 400 + 225 \][/tex]
[tex]\[ = 625\][/tex]

So, the left side equals the right side [tex]\(625\)[/tex]. Therefore, point A satisfies the equation.

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

Substitute [tex]\(x = -7\)[/tex] and [tex]\(y = 24\)[/tex] into the equation:
[tex]\((-7)^2 + (24 - 12)^2 = 49 + 12^2 \] \[ = 49 + 144 \] \[ = 193\] The left side does not equal the right side \(625\)[/tex]. Therefore, point B does not satisfy the equation.

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

Substitute [tex]\(x = 0\)[/tex] and [tex]\(y = 13\)[/tex] into the equation:
[tex]\[0^2 + (13 - 12)^2 = 0 + 1^2 \][/tex]
[tex]\[ = 0 + 1 \][/tex]
[tex]\[ = 1\][/tex]

The left side does not equal the right side [tex]\(625\)[/tex]. Therefore, point C does not satisfy the equation.

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

Substitute [tex]\(x = -25\)[/tex] and [tex]\(y = -13\)[/tex] into the equation:
[tex]\((-25)^2 + (-13 - 12)^2 = 625 + (-25)^2 \] \[ = 625 + 625 \] \[ = 1250\] The left side does not equal the right side \(625\)[/tex]. Therefore, point D does not satisfy the equation.

From the calculations above, only point A [tex]\((20, -3)\)[/tex] satisfies the circle equation [tex]\(x^2 + (y - 12)^2 = 25^2\)[/tex].

Therefore, the point that lies on the circle is: [tex]\((20, -3)\)[/tex], and the answer is A.
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