What is the radius of this equation after putting it into general circle form by completing the square? 5x^2+2x+5y^2-9y-3=0

Question

The general format of a circle equation: (x-h)^2+(y-k)^2=r^2

Answer 1

Adam:

 

First of all, you should be clear about what you want.  You DO NOT want the radius of the EQUATION (that verges on writing  nonsense), you want radius of the CIRCLE.  Please take pride in what you write.

 

Write

5x² + 2x as 5[x² + (2/5) x]

 

Now completer the square and adjust by subtracting the added constant (see below):

[x² + (2/5) x] = x² + 2 (1/5) x = x² + 2 (1/5) x + (1/5)² - (1/5)² = [ x + (1/5)]² - (1/25).

 

Do similar operations  for 5y² - 9y = 5[y² - (9/5)y]

y² - (9/5)y = y²  - 2(9/10)y = y² - 2(9/10)y + (9/10)² - (9/10)² = [y - (9/10)]² - (9/10)².

 

Combine everything.

 

5[x + (1/5)]² - 5(1/25) + 5[y - (9/10)]² - 5(9/10)² - 3= 0.

 

Divide throughout by 5.

 

[x + (1/5)]² - (1/25) + [y - (9/10)]² - (9/10)² - 3/5= 0.

 

This is the same as

 

[x + (1/5)]² + [y - (9/10)]² = (1/25) + (9/10)² + (3/5)

 

The term on the right equals (1/25) + (81/100) + (3/5)  = (4+81 + 60)/100 = 145/100 = [sqrt(145)/10]².

 

The above equation is of the form (x-h)²+(y-k)²=r², witj h = 1/5, k = -9/10 and r = [sqrt(145)/10].

 

So, the radius of the circle is sqrt(145)/10.

 

Having chastised  you (I am sorry!) in the beginning, I hope I have not made any arithmetical errors or typos. Of course, you have the right to scold me for them!  Take care.

 

Dattaprabhakar (Dr. G.0

 

 

Answer 2

x^2 + y ^2 +2/5*x -9/5*y -3/5=0

 

x^2+2/5*x +1/25-1/25+y^2-9/5*y+81/100-81/100-3/5=0

 

(x +1/5)^2 +(y-9/10)^2=3/5+81/100+1/25

 

                                    = (sqrt(29/20))^2

 

r= sqrt(29/20)