Dattaprabhakar G. answered • 08/29/14

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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

5x

^{2}+ 2x as 5[x^{2}+ (2/5) x]Now completer the square and adjust by subtracting the added constant (see below):

[x

^{2}+ (2/5) x] = x^{2}+ 2 (1/5) x = x^{2}+ 2 (1/5) x + (1/5)^{2}- (1/5)^{2}= [ x + (1/5)]^{2}- (1/25).Do similar operations for 5y

^{2}- 9y = 5[y^{2}- (9/5)y]y

^{2}- (9/5)y = y^{2}- 2(9/10)y = y^{2}- 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}=r^{2}, 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

Adam H.

08/30/14