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what is the standard form of the equation for this circle 4, -5 radius 5.5

Sagot :

olemakpadu
Standard form of equation of a circle:

(x - a)^2 + (y -b)^2  =  r ^2.

Where centre =  (a , b)
r  = radius.
Centre (4 , -5) = (a , b) ,  a = 4,  b = -5.
r = 5.5.

(x - 4)^2 + (y - -5)^2  = 5.5^2
(x - 4)^2 + (y +5)^2  = 5.5^2      Expand brackets, and use your calculator

x^2 - 8x + 16  +  y^2 +10y + 25 = 30.25 
x^2 - 8x +  y^2 +10y + 25 +16 - 30.25 = 0
x^2 - 8x +  y^2 +10y  - 10.25  = 0.   
x^2 + y^2 -8x +10y  - 10.25  = 0.       That's the equation of the circle.

or  10.25 =  10    /1/4  = 41/4

x^2 + y^2 -8x +10y  - 41/4  = 0    Multiply both sides of the equation by 4.

4x^2 + 4y^2 -32x +40y  - 41  = 0

Cheers.
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