help with calculus regarding circles

Question

please explain to me hoe to the get this equation of a circle, into standard form.

 

5x² -9x + 5y² -9y -2 = 0

Answer 1

Let me go step by step, 

Given, 5x² -9x + 5y² -9y -2 = 0

Divide by 5 and add 2 on both sides,  x² - 9x/5 + y² - 9y/5 = 2

Make perfect square,  

  {(x - 9/10)² - (9/10)²} + {(y - 9/10)² - (9/10)²} =  2

Add (9/10)² + (9/10)² on both side, 

  (x - 9/10)² +  (y - 9/10)² = 81/100 + 81/100 + 2 = 362/100

Therefore, (x - 9/10)² + (y - 9/10)² = (√(362) / 10)²

 

Answer 2

Hi Anabel:

 

The standard form of a circle is (x-h)² + (y-k)²  = r²     Where the center of the circle is at (h,k) and the radius of the circle is r.

 

Notice that if you expand the standard form of the above circle by performing the squaring operations you get:

 

x² -2xh + h² + y²  -2yk + k² = r²

 

Now this looks somewhat similar to your problem except for that coefficient of 5 on both the squared terms.  But this can easily be removed by dividing the entire equation by 5 and now, though you have some nasty fractions, your problem is beginning to look very much like the expanded version of the standard form.

 

Do you remember way back in Algebra I when you completed the square of a quadratic equation?  Well now you need to do this with both the x terms and the y terms and viola I'll bet you'll see the standard form emerge.

 

Yes, it will have some fractions and the center may be at (.9,.9) and the radius may have a radical but then again they're just numbers---right?