SOLUTION: find the center and the radius of the circle: x2 + y2 - 12x-16y-21 =0

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Question 61071: find the center and the radius of the circle:
x2 + y2 - 12x-16y-21 =0
Found 2 solutions by joyofmath, Earlsdon:
Answer by joyofmath(189) About Me  (Show Source):
You can put this solution on YOUR website!
find the center and the radius of the circle:
x%5E2+%2B+y%5E2+-+12x-16y-21+=+0
Add 36 to both sides: x%5E2-12x%2B36+%2B+y%5E2+-+12x-16y-21+=+36
Rewrite x-expression: %28x-6%29%5E2+%2B+y%5E2-16y-21+=+36
Add 64 to both sides: %28x-6%29%5E2+%2B+y%5E2-16y%2B64+-21+=+36%2B64=100
Rewrite y-expression: %28x-6%29%5E2+%2B+%28y-8%29%5E2+-+21+=+100
Add 21 to both sides: %28x-6%29%5E2+%2B+%28y-8%29%5E2+=+121
The center of the circle is (6,8).
The radius = sqrt%28121%29+=+11.

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Find the centre and radius of the circle:
x%5E2+%2B+y%5E2+-+12x+-+16y+-+21+=+0 First, add 21 to both sides.
x%5E2+%2B+y%5E2+-+12x+-+16y+=+21 Now group the x-terms together and the y-terms together.
%28x%5E2+-+12x%29+%2B+%28y%5E2+-+16y%29+=+21 Next, complete the square in x by adding the square of half the x-coefficient to both sides of the equation. %28-12%2F2%29%5E2+=+36, and the same for the y %28-16%2F2%29%5E2+=+64
%28x%5E2+-+12x+%2B+36%29+%2B+%28y%5E2+-+16y+%2B+64%29+=+21+%2B+36+%2B+64 Simplify.
%28x%5E2+-+12x+%2B+36%29+%2B+%28y%5E2+-+16y+%2B+64%29+=+121 Factor the x-group and the y-group.
%28x+-+6%29%5E2+%2B+%28y+-+8%29%5E2+=+121 Take the square root of 121 = 11
%28x+-+6%29%5E2+%2B+%28y+-+8%29%5E2+=+%2811%29%5E2 Compare with the standard form for a circle with centre at (h, k) and radius, r.
%28x+-+h%29%5E2+%2B+%28y+-+k%29%5E2+=+r%5E2
The circle centre is at (6, 8) and the radius is 11