Earthquakes release seismic waves that occur in concentric circles from the epicenter of the earthquake. Suppose a seismograph station determines the epicenter of an earthquake is located 9 kilometers from the station. If the epicenter is located at the origin, write the equation for the circular wave that passes through the station.
A. x2 + y2 = 81
B. x2 + y2 = 9
C. (x - 9)2 + (y - 9)2 = 0
D. (x + 9)2 + (y + 9)2 = 0

Question

Earthquakes release seismic waves that occur in concentric circles from the epicenter of the earthquake. Suppose a seismograph station determines the epicenter of an earthquake is located 9 kilometers from the station. If the epicenter is located at the origin, write the equation for the circular wave that passes through the station.
A. x2 + y2 = 81
B. x2 + y2 = 9
C. (x - 9)2 + (y - 9)2 = 0
D. (x + 9)2 + (y + 9)2 = 0

jamesj · Answer

Which of these equations is the equation of circle of radius 9 centered at the origin?

jamesj · Answer

Hint: the equation of a circle centered at $ (a,b) $ and of radius $ r $ is
$$ (x-a)^2 + (y-b)^2 = r^2 $$

anonymous · Answer

James <3 can you help me after by any chance?

jamesj · Answer

To make this as explicit as I can for you without giving the answer: if the circle is centered at the origin, then
$$ (a,b) = (0,0) $$
If the radius is 9 km, then $ r = 9 $.  Now, using those values of the variables and the general form of the circle I wrote above, what is the equation of a circle, centered at the origin and of radius 9?

jamesj · Answer

$$ (x-a)^2 - (y-b)^2 = r^2 $$  Now fill in the values of the variables, $a, b$ and $ r $.

jamesj · Answer

**correction +
$$ (x-a)^2 + (y-b)^2 = r^2 $$

anonymous · Answer

Okay so, 
Take the epicentre as the centre point of the circle (Since earthquake waves are circular.)
Then, since the epicentre is 9 km away, let the station be any point on the circle. So,
radius= 9^2 = 81
So, equation of the circle: x^2 + y^2 = 81 is your answer.|dw:1328818229713:dw|