The center of the circle is (-3,6)
The radius of the circle is [tex]\sqrt{34}[/tex] units
Equation of the circle is [tex]x^{2}[/tex] + [tex]y^{2}[/tex] +6x -6y + 11 = 0
A circle is the locus of points equidistant from a given point called the center of the circle.
Analysis:
Center of the circle is the mid point of the coordinates of the diameter P(-8,3) and Q(2,9).
x = x1+x2/2 = -8+2/2 = -6/2 = -3
y = y1+y2/2 = 3+9/2 = 12/2 = 6
coordinates of center are (-3,6)
To find the radius, first find the diameter of the circle
D = [tex]\sqrt{(x2-x1)^{2} + (y2-y1)^{2} }[/tex]
D = [tex]\sqrt{(2--8)^{2} + (9-3)^{2} }[/tex] = 2[tex]\sqrt{34}[/tex] units
Radius, R = D/2 = 2[tex]\sqrt{34}[/tex]/2 = [tex]\sqrt{34}[/tex] units
Equation of circle with centers a and b is:
[tex](x-a)^{2}[/tex] + [tex](y-b)^{2}[/tex] = [tex]r^{2}[/tex]
where a = -3, b = 6
[tex](x--3)^{2}[/tex] + [tex](y-6)^{2}[/tex] = [tex]\sqrt{34} ^{2}[/tex]
[tex](x+3)^{2}[/tex] + [tex](y-6)^{2}[/tex] = 34
[tex]x^{2}[/tex] +6x + 9 + [tex]y^{2}[/tex] -12y +36 = 34
[tex]x^{2}[/tex] + [tex]y^{2}[/tex] + 6x -12y + 11 = 0
In conclusion,
The center of the circle is (-3,6)
The radius of the circle is [tex]\sqrt{34}[/tex] units
Equation of the circle is [tex]x^{2}[/tex] + [tex]y^{2}[/tex] +6x -6y + 11 = 0
Learn more about equation of circle: brainly.com/question/1506955
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