pinksosa pinksosa
  • 15-02-2017
  • Mathematics
contestada

Write the equation of the circle with center (−3, 2) and (6, 4) a point on the circle.

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sheheryar91
sheheryar91 sheheryar91
  • 15-02-2017
radius=sqrt(2^2+9^2)
r=9.22
equation
(x+3)^2+(y-2)^2=9.22^2
(x+3)^2+(y-2)^2=85
Answer Link
apologiabiology
apologiabiology apologiabiology
  • 15-02-2017
[tex](x-h)^2+(y-k)^2=r^2[/tex]
center is (h,k) and radius is r

so

given center=(-3,2)=(h,k)
[tex](x-(-3))^2+(y-2)^2=r^2[/tex]
[tex](x+3)^2+(y-2)^2=r^2[/tex]
find r^2
ok, (6,4) is a point
sub 6 for x and 4 for y and evaluate

[tex](6+3)^2+(4-2)^2=r^2[/tex]
[tex](9)^2+(2)^2=r^2[/tex]
[tex]81+4=r^2[/tex]
[tex]85=r^2[/tex]


the equation is
[tex](x+3)^2+(y-2)^2=85[/tex]
Answer Link

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