SOLUTION: write the equation in standard form: xsquared+ysquared-6x-4y+13=36 sorry that i cant find a small 2 for the squared. i hope this makes sense. please help me figure this out. t

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: write the equation in standard form: xsquared+ysquared-6x-4y+13=36 sorry that i cant find a small 2 for the squared. i hope this makes sense. please help me figure this out. t      Log On


   



Question 47791: write the equation in standard form:
xsquared+ysquared-6x-4y+13=36
sorry that i cant find a small 2 for the squared. i hope this makes sense. please help me figure this out. thanks so much!

Found 2 solutions by pizza, stanbon:
Answer by pizza(14) About Me  (Show Source):
You can put this solution on YOUR website!
The first thing that needs clarification is the term 'standard form'.
The equation happens to be in standard form already.
However, I deduce from the title, that by standard form, you want to rearrange the equation such that it is evidently a conic section. The standard form of a conic section is as follows:
+%28%28x-a%29%5E2%29%2Fc+%2B+%28%28y-b%29%5E2%29%2Fd+=+r%5E2+
This implies that you have to do some factorisation.
I assume you are familiar with completing the square.
If you are, then you will know that
+x%5E2+-+6x+=+%28x-3%29%5E2+-+9+
And
+y%5E2+-+4x+=+%28y-2%29%5E2+-+4+
Then putting this into the original equation, we have
+%28x-3%29%5E2+%2B+%28y-2%29%5E2+=+36+=+6%5E2+
which is in standard form.
Hooray!
When c = d = 1, we have an equation for a circle, otherwise it is an equation of an ellipse. Circles and ellipses are conic sections.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
x^2+y^2-6x-4y+13=36
Complete the square on the x terms and on the y terms separately to get:
x^2-6x+9 +y^2-4y+4 =36-13+9+4
(x-3)^2 + (y-2)^2 = 36
This is the equation for a circle with center
at (3,2) and radius of 6
Cheers,
Stan H.