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kobusy [5.1K]
3 years ago
10

A circle has a diameter with endpoints at 3 – 5i and –8 + 2i. What is the center of the circle?

Mathematics
2 answers:
mrs_skeptik [129]3 years ago
4 0

Option B.) (-5/2)(-3/2)i

bagirrra123 [75]3 years ago
3 0
So we got the real axis and the imaginary axis
we just need to find the average of the 2 points

remember
midpoint of (x1,y1) and (x2,y2) is

((x1+x2)/2,(y1+y2)/2)
so

average of 3 and -8 is -5/2
average of -5i and 2i is -3/2i

center is -5/2-3/2i
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By ii) \displaystyle{b^2-4ac=0.

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          \displaystyle{b^2-4ac=0
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          \displaystyle{b^2+4(25-5b)=0
          \displaystyle{b^2-20b+100=0
          \displaystyle{(b-10)^2=0
          \displaystyle{b=10 


Finally we find c: c=25-5b=25-50=-25

Thus the function is        \displaystyle{f(x)=-x^2+10x-25


Remark: It is also possible to solve the problem by considering the form

f(x)=-1(x-5)^2 directly.

In general, if a quadratic function has leading coefficient a, and has a root r of multiplicity 2, then its form is f(x)=a(x-r)^2
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