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# How do I get the circle equation x² + y² = 10x + 5y -25 in terms of y?

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I don't understand what to do with the x^2 + y^2

Guest Dec 16, 2014

#1
+83944
+8

#### x² + y² = 10x + 5y -25     subtract 10x, 5y from both sides, so we have

x^2 - 10x + y^2 - 5y  = -25     complete the square on x and y..... so we have

x^2 - 10x + 25  + y^2 - 5y + 25/4  = -25 +25 + 25/4     factor and simplify

(x - 5)^2 + ( y - 5/2)^2  = 25/4

So, this is a circle whose center is ( 5, 5/2) and whose radius is 5/2

Now if we wanted this in terms of y.....first subtract (x - 5)^2 from both sides

( y - 5/2)^2 = 25/4 - (x - 5)^2    take the positive and negative square root of both sides

y -5/2  = ±√[25/4 - (x - 5)^2]    add 5/2 to both sides

y = ±√[25/4 - (x - 5)^2] + 5/2

And that's everything in terms of 'y"

CPhill  Dec 16, 2014
Sort:

#1
+83944
+8

#### x² + y² = 10x + 5y -25     subtract 10x, 5y from both sides, so we have

x^2 - 10x + y^2 - 5y  = -25     complete the square on x and y..... so we have

x^2 - 10x + 25  + y^2 - 5y + 25/4  = -25 +25 + 25/4     factor and simplify

(x - 5)^2 + ( y - 5/2)^2  = 25/4

So, this is a circle whose center is ( 5, 5/2) and whose radius is 5/2

Now if we wanted this in terms of y.....first subtract (x - 5)^2 from both sides

( y - 5/2)^2 = 25/4 - (x - 5)^2    take the positive and negative square root of both sides

y -5/2  = ±√[25/4 - (x - 5)^2]    add 5/2 to both sides

y = ±√[25/4 - (x - 5)^2] + 5/2

And that's everything in terms of 'y"

CPhill  Dec 16, 2014

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