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  • 02-08-2018
  • Mathematics
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Identify the equation of the circle that has its center at (9, 12) and passes through the origin.

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sarajevo
sarajevo sarajevo
  • 02-08-2018

The circle has center at (9, 12) and passes through the origin.

Equation of the circle in center and radius form is given by

[tex] (x-h)^2+(y-k)^2=r^2 [/tex]

where r is the radius and center at (h,k)

Now substitute the value of the center we get

[tex] (x-9)^2+(y-12)^2=r^2 [/tex]

As it passes through the origin so we can write

[tex] (0-9)^2+(0-12)^2=r^2\\
\\
81+144=r^2\\
\\
225=r^2\\
\\
r=15\\ [/tex]

Hence the equation of the circle is

[tex] (x-9)^2+(y-12)^2=15^2 [/tex]

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