SOLUTION: Identify the conic(cirlce, ellipse, hyperbola, ellipse. Write each of the following in standard form. Find the information listed for each problem. Graph the equation. 1) x^2+y^

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Identify the conic(cirlce, ellipse, hyperbola, ellipse. Write each of the following in standard form. Find the information listed for each problem. Graph the equation. 1) x^2+y^      Log On


   

Question 83616: Identify the conic(cirlce, ellipse, hyperbola, ellipse. Write each of the following in standard form. Find the information listed for each problem. Graph the equation.
1) x^2+y^2-16x+12y=0
What is it?_________
Center_________
Radius__________
Standard Form_________
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
1+%2A+x+%5E+2+%2B+y+%5E+2+-+1+6+%2A+x+%2B+1+2+%2A+y++=0 Start with the given conic



%28+x+%5E+2+-+1+6+%2A+x+%29+%2B+%28+y+%5E+2+%2B+1+2+%2A+y+%29+=+0+++++ Group like terms




Now we must complete the individual squares inside the parenthesis:



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For instance to complete 1+%28+x+%5E+2+-+1+6+%2A+x+%29+, take half of -16 and square it (ie %28-16%2F2%29%5E2=%28-8%29%5E2=64 to get 64. Now add 64 inside the parenthesis like this:
%28+x+%5E+2+-+1+6+%2A+x+%2B+6+4+%29+
Since you really added 1%2A64 to the entire left side, we must add 1%2A64 to the right side also. Now lets complete +%28+y+%5E+2+%2B+1+2+%2A+y+%29

In order to complete %28+y+%5E+2+%2B+1+2+%2A+y+%29, take half of 12 and square it (ie %2812%2F2%29%5E2=%286%29%5E2=36 to get 36. Now add 36 inside the parenthesis like this:
+%28+y+%5E+2+%2B+1+2+%2A+y+%2B+3+6+%29
Since you really added 1%2A36 to the entire left side, we must add 1%2A36 to the right side also.

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Complete the individual squares by taking half of the 2nd coefficient and squaring it (remember to add to both sides).

+%28+x+%5E+2+-+1+6+%2A+x+%2B+6+4+%29+%2B+%28+y+%5E+2+%2B+1+2+%2A+y+%2B+3+6+%29+=+1+0+0+ Combine like terms on the right side




+%28+x+-+8+%29+%5E+2+%2B++%28+y+%2B+6+%29+%5E+2+=+1+0+0+ Factor the individual groups on the left side.

So the equation is now in standard form %28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2
where (h,k) is the center and r is the radius

So the equation is a circle with a radius of 10 and a center of (8,-6)

Here's the graph of %28+x+-+8+%29+%5E+2+%2B+%28+y+%2B+6+%29+%5E+2+=+1+0+0+



What is it? Circle
Center: (8,-6)
Radius: 10
Standard Form: %28+x+-+8+%29+%5E+2+%2B+%28+y+%2B+6+%29+%5E+2+=+1+0+0+