x^2+y^2-3x-4y+4=0 How is this broken down for a circle
first seperate your variables from each other so put all your x's and y's next to each other and then subtract 4
\[(x^2-3x)+(y^2-4y)=-4\]
then complete the squares for x and y
do you know how to do this
I am not real sure how to do this.
Completing the Square method will be used here..
completing the square happens when you have an equation such as \[x^2+bx=0\] and you want to make it into a perfect square. to do this you take the coefficient b and divide it by two. after dividing it, square it and add it to both sides of the equation. the end product looks like this \[x^2+bx+\frac{b^2}{2}=\frac{b^2}{2}\]
there is a mistake in the end... the form should be this \[x^2+bx+(\frac{b}{2})^2=(\frac{b}{2})^2\]
the left side can be factored into a perfect square \[(x+\frac{b}{2})^2\]=(\frac{b}{2})^2\]
so if you do that with both x and y what do you gte
confused
Thanks for the help!!!
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