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Sagot :
To find the equation of the circle with the line segment [tex]\( PQ \)[/tex] as its diameter, follow these steps:
1. Find the midpoint of segment [tex]\( PQ \)[/tex]:
The midpoint [tex]\( M \)[/tex] of segment [tex]\( PQ \)[/tex] can be computed using the midpoint formula:
[tex]\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right). \][/tex]
For [tex]\( P = (3, 1) \)[/tex] and [tex]\( Q = (-3, -7) \)[/tex], the coordinates of the midpoint are:
[tex]\[ M = \left( \frac{3 + (-3)}{2}, \frac{1 + (-7)}{2} \right) = \left( \frac{0}{2}, \frac{-6}{2} \right) = (0, -3). \][/tex]
So, the center of the circle is [tex]\( (0, -3) \)[/tex].
2. Calculate the radius of the circle:
The radius [tex]\( r \)[/tex] of the circle is half the length of segment [tex]\( PQ \)[/tex]:
[tex]\[ r = \frac{1}{2} \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. \][/tex]
Using the coordinates of [tex]\( P \)[/tex] and [tex]\( Q \)[/tex], we compute:
[tex]\[ r = \frac{1}{2} \sqrt{((-3) - 3)^2 + ((-7) - 1)^2} = \frac{1}{2} \sqrt{(-6)^2 + (-8)^2} = \frac{1}{2} \sqrt{36 + 64} = \frac{1}{2} \sqrt{100} = \frac{1}{2} \times 10 = 5. \][/tex]
3. Write the standard form of the circle's equation:
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 the values found:
[tex]\[ h = 0, \quad k = -3, \quad r = 5, \][/tex]
thus, the equation of the circle is:
[tex]\[ (x - 0)^2 + (y + 3)^2 = 5^2, \][/tex]
which simplifies to:
[tex]\[ x^2 + (y + 3)^2 = 25. \][/tex]
So, the equation of the circle is:
\[
(x - 0)^2 + (y + 3)^2 = 25.
\
1. Find the midpoint of segment [tex]\( PQ \)[/tex]:
The midpoint [tex]\( M \)[/tex] of segment [tex]\( PQ \)[/tex] can be computed using the midpoint formula:
[tex]\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right). \][/tex]
For [tex]\( P = (3, 1) \)[/tex] and [tex]\( Q = (-3, -7) \)[/tex], the coordinates of the midpoint are:
[tex]\[ M = \left( \frac{3 + (-3)}{2}, \frac{1 + (-7)}{2} \right) = \left( \frac{0}{2}, \frac{-6}{2} \right) = (0, -3). \][/tex]
So, the center of the circle is [tex]\( (0, -3) \)[/tex].
2. Calculate the radius of the circle:
The radius [tex]\( r \)[/tex] of the circle is half the length of segment [tex]\( PQ \)[/tex]:
[tex]\[ r = \frac{1}{2} \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. \][/tex]
Using the coordinates of [tex]\( P \)[/tex] and [tex]\( Q \)[/tex], we compute:
[tex]\[ r = \frac{1}{2} \sqrt{((-3) - 3)^2 + ((-7) - 1)^2} = \frac{1}{2} \sqrt{(-6)^2 + (-8)^2} = \frac{1}{2} \sqrt{36 + 64} = \frac{1}{2} \sqrt{100} = \frac{1}{2} \times 10 = 5. \][/tex]
3. Write the standard form of the circle's equation:
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 the values found:
[tex]\[ h = 0, \quad k = -3, \quad r = 5, \][/tex]
thus, the equation of the circle is:
[tex]\[ (x - 0)^2 + (y + 3)^2 = 5^2, \][/tex]
which simplifies to:
[tex]\[ x^2 + (y + 3)^2 = 25. \][/tex]
So, the equation of the circle is:
\[
(x - 0)^2 + (y + 3)^2 = 25.
\
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