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Sagot :
To determine which point lies on the circle given by the equation [tex]\((x-3)^2 + (y+4)^2 = 6^2\)[/tex], let's analyze the situation step by step.
1. Understand the Circle Equation:
- The standard form of a circle's equation is [tex]\((x-h)^2 + (y-k)^2 = r^2\)[/tex], where [tex]\((h, k)\)[/tex] is the center of the circle and [tex]\(r\)[/tex] is the radius.
- From the equation [tex]\((x-3)^2 + (y+4)^2 = 6^2\)[/tex], we can identify:
- The center [tex]\((h, k)\)[/tex] of the circle is [tex]\((3, -4)\)[/tex].
- The radius [tex]\(r\)[/tex] of the circle is [tex]\(6\)[/tex].
2. Calculate the Distances from the Center to Each Point:
- The distance [tex]\(d\)[/tex] from the center [tex]\((3, -4)\)[/tex] to a point [tex]\((x, y)\)[/tex] is given by the formula:
[tex]\[ d = \sqrt{(x - 3)^2 + (y + 4)^2} \][/tex]
3. Check Each Point:
- For point [tex]\((9, -2)\)[/tex]:
[tex]\[ d = \sqrt{(9 - 3)^2 + (-2 + 4)^2} = \sqrt{6^2 + 2^2} = \sqrt{36 + 4} = \sqrt{40} \approx 6.32 \][/tex]
- The distance is approximately 6.32, not equal to the radius 6.
- For point [tex]\((0, 11)\)[/tex]:
[tex]\[ d = \sqrt{(0 - 3)^2 + (11 + 4)^2} = \sqrt{(-3)^2 + 15^2} = \sqrt{9 + 225} = \sqrt{234} \approx 15.3 \][/tex]
- The distance is approximately 15.3, not equal to the radius 6.
- For point [tex]\((3, 10)\)[/tex]:
[tex]\[ d = \sqrt{(3 - 3)^2 + (10 + 4)^2} = \sqrt{0^2 + 14^2} = \sqrt{0 + 196} = \sqrt{196} = 14 \][/tex]
- The distance is 14, not equal to the radius 6.
- For point [tex]\(( - 9, 4\)[/tex]):
[tex]\[ d = \sqrt{(-9 - 3)^2 + (4 + 4)^2} = \sqrt{(-12)^2 + 8^2} = \sqrt{144 + 64} = \sqrt{208} \approx 14.42 \][/tex]
- The distance is approximately 14.42, not equal to the radius 6.
- For point [tex]\((-3, -4)\)[/tex]:
[tex]\[ d = \sqrt{(-3 - 3)^2 + (-4 + 4)^2} = \sqrt{(-6)^2 + 0^2} = \sqrt{36 + 0} = \sqrt{36} = 6 \][/tex]
- The distance is exactly 6, which is equal to the radius of the circle.
4. Conclusion:
- The only point whose distance from the center [tex]\((3, -4)\)[/tex] is exactly equal to the radius [tex]\(6\)[/tex] is [tex]\((-3, -4)\)[/tex].
Therefore, the correct answer is:
[tex]\[ \boxed{E. (-3, -4)} \][/tex]
1. Understand the Circle Equation:
- The standard form of a circle's equation is [tex]\((x-h)^2 + (y-k)^2 = r^2\)[/tex], where [tex]\((h, k)\)[/tex] is the center of the circle and [tex]\(r\)[/tex] is the radius.
- From the equation [tex]\((x-3)^2 + (y+4)^2 = 6^2\)[/tex], we can identify:
- The center [tex]\((h, k)\)[/tex] of the circle is [tex]\((3, -4)\)[/tex].
- The radius [tex]\(r\)[/tex] of the circle is [tex]\(6\)[/tex].
2. Calculate the Distances from the Center to Each Point:
- The distance [tex]\(d\)[/tex] from the center [tex]\((3, -4)\)[/tex] to a point [tex]\((x, y)\)[/tex] is given by the formula:
[tex]\[ d = \sqrt{(x - 3)^2 + (y + 4)^2} \][/tex]
3. Check Each Point:
- For point [tex]\((9, -2)\)[/tex]:
[tex]\[ d = \sqrt{(9 - 3)^2 + (-2 + 4)^2} = \sqrt{6^2 + 2^2} = \sqrt{36 + 4} = \sqrt{40} \approx 6.32 \][/tex]
- The distance is approximately 6.32, not equal to the radius 6.
- For point [tex]\((0, 11)\)[/tex]:
[tex]\[ d = \sqrt{(0 - 3)^2 + (11 + 4)^2} = \sqrt{(-3)^2 + 15^2} = \sqrt{9 + 225} = \sqrt{234} \approx 15.3 \][/tex]
- The distance is approximately 15.3, not equal to the radius 6.
- For point [tex]\((3, 10)\)[/tex]:
[tex]\[ d = \sqrt{(3 - 3)^2 + (10 + 4)^2} = \sqrt{0^2 + 14^2} = \sqrt{0 + 196} = \sqrt{196} = 14 \][/tex]
- The distance is 14, not equal to the radius 6.
- For point [tex]\(( - 9, 4\)[/tex]):
[tex]\[ d = \sqrt{(-9 - 3)^2 + (4 + 4)^2} = \sqrt{(-12)^2 + 8^2} = \sqrt{144 + 64} = \sqrt{208} \approx 14.42 \][/tex]
- The distance is approximately 14.42, not equal to the radius 6.
- For point [tex]\((-3, -4)\)[/tex]:
[tex]\[ d = \sqrt{(-3 - 3)^2 + (-4 + 4)^2} = \sqrt{(-6)^2 + 0^2} = \sqrt{36 + 0} = \sqrt{36} = 6 \][/tex]
- The distance is exactly 6, which is equal to the radius of the circle.
4. Conclusion:
- The only point whose distance from the center [tex]\((3, -4)\)[/tex] is exactly equal to the radius [tex]\(6\)[/tex] is [tex]\((-3, -4)\)[/tex].
Therefore, the correct answer is:
[tex]\[ \boxed{E. (-3, -4)} \][/tex]
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