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Identify the equation of the circle that has its center at [tex]$(-16,30)$[/tex] and passes through the origin.

A. [tex]$(x-16)^2+(y+30)^2=1256$[/tex]

B. [tex][tex]$(x+16)^2+(y-30)^2=1256$[/tex][/tex]

C. [tex]$(x-16)^2+(y+30)^2=1156$[/tex]

D. [tex]$(x+16)^2+(y-30)^2=1156$[/tex]


Sagot :

To identify the equation of the circle that has its center at [tex]\((-16,30)\)[/tex] and passes through the origin [tex]\((0, 0)\)[/tex], we need to follow these steps:

1. Determine the radius of the circle:
- The radius [tex]\( r \)[/tex] of the circle can be found by calculating the distance from the center to the point through which it passes (the origin in this case).
- The distance formula for two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:
[tex]\[ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
- Here, the coordinates of the center are [tex]\((x_1, y_1) = (-16, 30)\)[/tex], and the coordinates of the origin are [tex]\((x_2, y_2) = (0, 0)\)[/tex].

[tex]\[ r = \sqrt{(0 - (-16))^2 + (0 - 30)^2} \][/tex]
[tex]\[ r = \sqrt{(16)^2 + (-30)^2} \][/tex]
[tex]\[ r = \sqrt{256 + 900} \][/tex]
[tex]\[ r = \sqrt{1156} \][/tex]
[tex]\[ r = 34 \][/tex]

2. Write the general equation of the circle:
- The standard form of the equation of a circle with center [tex]\((h, k)\)[/tex] and radius [tex]\(r\)[/tex] is:
[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]

- In our case, the center [tex]\((h, k) = (-16, 30)\)[/tex] and the radius [tex]\(r = 34\)[/tex].
- So, plugging in these values, the equation becomes:
[tex]\[ (x - (-16))^2 + (y - 30)^2 = 34^2 \][/tex]
[tex]\[ (x + 16)^2 + (y - 30)^2 = 1156 \][/tex]

3. Compare with the options provided:
- The given options are:
- A: [tex]\((x - 16)^2 + (y + 30)^2 = 1256\)[/tex]
- B: [tex]\((x + 16)^2 + (y - 30)^2 = 1256\)[/tex]
- C: [tex]\((x - 16)^2 + (y + 30)^2 = 1156\)[/tex]
- D: [tex]\((x + 16)^2 + (y - 30)^2 = 1156\)[/tex]

- The correct form of the equation we derived is:
[tex]\[ (x + 16)^2 + (y - 30)^2 = 1156 \][/tex]

- Comparing this with the options, we see that option D matches perfectly.

Therefore, the equation of the circle that has its center at [tex]\((-16,30)\)[/tex] and passes through the origin is:

[tex]\[ \boxed{(x + 16)^2 + (y - 30)^2 = 1156} \][/tex]