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Sagot :
To determine the equation of the circle in standard form, given that points [tex]\(P(-10,10)\)[/tex] and [tex]\(Q(6,-2)\)[/tex] give the diameter of the circle, we must follow these steps:
1. Find the center of the circle:
Since [tex]\(P\)[/tex] and [tex]\(Q\)[/tex] are the endpoints of the diameter, the center of the circle is the midpoint of the segment [tex]\(PQ\)[/tex].
The midpoint [tex]\((h, k)\)[/tex] of the segment joining points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is calculated as:
[tex]\[ h = \frac{x_1 + x_2}{2}, \quad k = \frac{y_1 + y_2}{2} \][/tex]
Plugging in the coordinates of [tex]\(P\)[/tex] and [tex]\(Q\)[/tex]:
[tex]\[ x_1 = -10, \quad y_1 = 10 \][/tex]
[tex]\[ x_2 = 6, \quad y_2 = -2 \][/tex]
Therefore:
[tex]\[ h = \frac{-10 + 6}{2} = \frac{-4}{2} = -2 \][/tex]
[tex]\[ k = \frac{10 - 2}{2} = \frac{8}{2} = 4 \][/tex]
Hence, the center [tex]\((h, k)\)[/tex] of the circle is [tex]\((-2, 4)\)[/tex].
2. Find the radius of the circle:
The radius [tex]\(r\)[/tex] is half the length of the diameter [tex]\(PQ\)[/tex]. We first need to calculate the distance [tex]\(PQ\)[/tex] using the distance formula:
[tex]\[ PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Substituting the coordinates:
[tex]\[ PQ = \sqrt{(6 - (-10))^2 + (-2 - 10)^2} = \sqrt{(6 + 10)^2 + (-2 - 10)^2} \][/tex]
[tex]\[ PQ = \sqrt{16^2 + (-12)^2} = \sqrt{256 + 144} = \sqrt{400} = 20 \][/tex]
Thus, the radius [tex]\(r\)[/tex] is:
[tex]\[ r = \frac{PQ}{2} = \frac{20}{2} = 10 \][/tex]
3. Write the equation of the circle in standard form:
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]
Substituting [tex]\(h = -2\)[/tex], [tex]\(k = 4\)[/tex], and [tex]\(r = 10\)[/tex]:
[tex]\[ (x - (-2))^2 + (y - 4)^2 = 10^2 \][/tex]
[tex]\[ (x + 2)^2 + (y - 4)^2 = 100 \][/tex]
Therefore, the equation of the circle in standard form is:
[tex]\[ (x + 2)^2 + (y - 4)^2 = 100 \][/tex]
So, the correct answer is:
[tex]\[ \boxed{(x + 2)^2 + (y - 4)^2 = 100} \][/tex]
1. Find the center of the circle:
Since [tex]\(P\)[/tex] and [tex]\(Q\)[/tex] are the endpoints of the diameter, the center of the circle is the midpoint of the segment [tex]\(PQ\)[/tex].
The midpoint [tex]\((h, k)\)[/tex] of the segment joining points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is calculated as:
[tex]\[ h = \frac{x_1 + x_2}{2}, \quad k = \frac{y_1 + y_2}{2} \][/tex]
Plugging in the coordinates of [tex]\(P\)[/tex] and [tex]\(Q\)[/tex]:
[tex]\[ x_1 = -10, \quad y_1 = 10 \][/tex]
[tex]\[ x_2 = 6, \quad y_2 = -2 \][/tex]
Therefore:
[tex]\[ h = \frac{-10 + 6}{2} = \frac{-4}{2} = -2 \][/tex]
[tex]\[ k = \frac{10 - 2}{2} = \frac{8}{2} = 4 \][/tex]
Hence, the center [tex]\((h, k)\)[/tex] of the circle is [tex]\((-2, 4)\)[/tex].
2. Find the radius of the circle:
The radius [tex]\(r\)[/tex] is half the length of the diameter [tex]\(PQ\)[/tex]. We first need to calculate the distance [tex]\(PQ\)[/tex] using the distance formula:
[tex]\[ PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Substituting the coordinates:
[tex]\[ PQ = \sqrt{(6 - (-10))^2 + (-2 - 10)^2} = \sqrt{(6 + 10)^2 + (-2 - 10)^2} \][/tex]
[tex]\[ PQ = \sqrt{16^2 + (-12)^2} = \sqrt{256 + 144} = \sqrt{400} = 20 \][/tex]
Thus, the radius [tex]\(r\)[/tex] is:
[tex]\[ r = \frac{PQ}{2} = \frac{20}{2} = 10 \][/tex]
3. Write the equation of the circle in standard form:
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]
Substituting [tex]\(h = -2\)[/tex], [tex]\(k = 4\)[/tex], and [tex]\(r = 10\)[/tex]:
[tex]\[ (x - (-2))^2 + (y - 4)^2 = 10^2 \][/tex]
[tex]\[ (x + 2)^2 + (y - 4)^2 = 100 \][/tex]
Therefore, the equation of the circle in standard form is:
[tex]\[ (x + 2)^2 + (y - 4)^2 = 100 \][/tex]
So, the correct answer is:
[tex]\[ \boxed{(x + 2)^2 + (y - 4)^2 = 100} \][/tex]
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