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If [tex]$P=(3,1)$[/tex] and [tex]$Q=(-3,-7)$[/tex], find the equation of the circle that has segment [tex][tex]$PQ$[/tex][/tex] as a diameter.

[tex]$(x - [?])^2 + (y - [?])^2 = [?]$[/tex]


Sagot :

To find the equation of the circle that has a segment [tex]\( PQ \)[/tex] as its diameter, let's go through the steps systematically:

1. Identify Points [tex]\( P \)[/tex] and [tex]\( Q \)[/tex]:
- Point [tex]\( P \)[/tex] is given as [tex]\( (3, 1) \)[/tex].
- Point [tex]\( Q \)[/tex] is given as [tex]\( (-3, -7) \)[/tex].

2. Find the Midpoint of [tex]\( PQ \)[/tex]:
The midpoint of segment [tex]\( PQ \)[/tex] will be the center of the circle. The midpoint [tex]\( (C_x, C_y) \)[/tex] can be calculated using the midpoint formula:
[tex]\[ C_x = \frac{x_1 + x_2}{2}, \qquad C_y = \frac{y_1 + y_2}{2} \][/tex]
Here, [tex]\( (x_1, y_1) = (3, 1) \)[/tex] and [tex]\( (x_2, y_2) = (-3, -7) \)[/tex].

[tex]\[ C_x = \frac{3 + (-3)}{2} = \frac{0}{2} = 0 \][/tex]
[tex]\[ C_y = \frac{1 + (-7)}{2} = \frac{-6}{2} = -3 \][/tex]
Therefore, the center of the circle is at [tex]\( (0, -3) \)[/tex].

3. Calculate the Radius of the Circle:
The radius is half the length of segment [tex]\( PQ \)[/tex]. First, calculate the distance between points [tex]\( P \)[/tex] and [tex]\( Q \)[/tex]:
[tex]\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
[tex]\[ \text{Distance} = \sqrt{(-3 - 3)^2 + (-7 - 1)^2} = \sqrt{(-6)^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \][/tex]
Since the diameter of the circle is 10, the radius [tex]\( r \)[/tex] is:
[tex]\[ r = \frac{10}{2} = 5 \][/tex]

4. Form the Equation of the Circle:
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 and [tex]\( r \)[/tex] is the radius. substituting [tex]\( h = 0 \)[/tex], [tex]\( k = -3 \)[/tex], and [tex]\( r = 5 \)[/tex]:
[tex]\[ (x - 0)^2 + (y - (-3))^2 = 5^2 \][/tex]
Simplifying this:
[tex]\[ x^2 + (y + 3)^2 = 25 \][/tex]

Thus, the equation of the circle that has the segment [tex]\( PQ \)[/tex] as its diameter is:
[tex]\[ (x - 0)^2 + (y - (-3))^2 = 25 \][/tex]
or more simply:
[tex]\[ (x - 0)^2 + (y + 3)^2 = 25 \][/tex]
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