SOLUTION: Can anyone tell me how to: Find an equation of the circle with center (4,-3) and radius 5. I need to express the final equation in the form x squared + y squared + Dx + Ey + F = 0

Question 35464This question is from textbook algebra for college students
: Can anyone tell me how to: Find an equation of the circle with center (4,-3) and radius 5. I need to express the final equation in the form x squared + y squared + Dx + Ey + F = 0 This question is from textbook algebra for college students

Found 3 solutions by Nate, rapaljer, stanbon:
Answer by Nate(3500)   (Show Source): You can put this solution on YOUR website!
the equation for a circle is:
(x-h)^2 + (y-k)^2 = r^2 where center is (h,k) and r is the radius
with center (4,-3) and radius 5
(x - 4)^2 + (y + 3)^2 = 25
x^2 - 8x + 16 + y^2 + 6y + 9 - 25 = 0
x^2 + y^2 - 8x + 6y = 0
Answer by rapaljer(4671)   (Show Source): You can put this solution on YOUR website!
The equation of a circle with radius r and center at (h, k) is

Therefore the equation is

Expanding:

Write it in standard form:
, which incidentally, passes through the origin.

R^2 at SCC
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
Can anyone tell me how to: Find an equation of the circle with center (4,-3) and radius 5. I need to express the final equation in the form x squared + y squared + Dx + Ey + F = 0
EQuation:
(x-4)^2 + (y+3)^2 = 5^2
x^2 + y^2 -8x + 6y +16 +9 - 25 = 0
x^2 + y^2 -8x + 6y+0=0
Cheers,
Stan H.

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