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A circle with center ( a, b ) and radius r has equation given as \[(x-a)^2+(y-b)^2=r^2\] We are given equation of circle as \[(x – 16)^2 + (y + 9)^2 = 4\] @julie001 Can you compare the two equations to find the center and radius?
Center: (16, -9); Radius: 4 Center: (-16, 9); Radius: 2 Center: (16, -9); Radius: 2 Center: (-16, 9); Radius: 4
im not sure
@sugargurl Please explain how did you get the answer?
@julie001 where you have doubt?
comparintg the equations
ANSWER IS THE 2ND CHOICE..:)
No sugargurl. Answer is NOT the second choice.
Work step by step \[(x-a)^2=(x-16)^2\] so a=16 Now you could find b and r, using the same procedure
YUO THE ANSWER IS THE 3RD CHOICE THANKS FOR CORRECTING ME @ MERTSJ
@ash2326 im still lost
Did you understand how I found a?
from the question ?
Yeah, standard equation of a circle is \[(x-a)^2+(y-b)^2=r^2\] and the equation given in the question is \[(x-16)^2+(y+9)^2=4\] We need to find the center (a, b) and radius r. So we compare the two equations \[(x-a)^2=(x-16)^2\] To make both sides equal a should be 16, so a=16 Now can you find b, by comparing \[(y-b)^2=(y+9)^2\]
so its Center: (16, -9); Radius: 2
Could you show me your work?
There is no work to show. We know that if the equation is written in the form: \[(x-h)^2+(y-k)^2=r^2\] then the center is (h,k) and the radius is r
i just plugged in the numbers with the equation
So write the given equation: \[(x-16)^2+(y-(-9))^2=2^2\] and it is immediately obvious that the center is (16,-9) and the radius is 2
@Mertsj is it Center: (16, -9); Radius: 2
and it is immediately obvious that the center is (16,-9) and the radius is 2