Oswaldo Riley

2023-03-20

How to graph ${x}^{2}+{y}^{2}-4x-2y-4=0$?

popunjati7z8a

Beginner2023-03-21Added 4 answers

You can determine that the equation represents a circle from this form.

To find its coordinates and radius you should transform it to form of:

$(x-a)}^{2}+{(y-b)}^{2}={r}^{2$ (1)

We start from the equation given:

${x}^{2}+{y}^{2}-4x-2y-4=0$

Now we can group terms with the same variable:

${x}^{2}-4x+{y}^{2}-2y-4=0$

Now we can complete the squares of variables:

${x}^{2}-4x{+4}+{y}^{2}-2y{+1}{-5}-4=0$

I added $5$ so I had to substract 5 to keep the equation balanced.

Now we can write the expressions with $x$ and $y$ as squares:

${(x-2)}^{2}+{(y-1)}^{2}-9=0$

Finally I can move $9$ to right side to get the form (1)

${(x-2)}^{2}+{(y-1)}^{2}=9$

From this equation I can read that the equation shows a circle with center in $C=(2;1)$ and radius $r=3$

To find its coordinates and radius you should transform it to form of:

$(x-a)}^{2}+{(y-b)}^{2}={r}^{2$ (1)

We start from the equation given:

${x}^{2}+{y}^{2}-4x-2y-4=0$

Now we can group terms with the same variable:

${x}^{2}-4x+{y}^{2}-2y-4=0$

Now we can complete the squares of variables:

${x}^{2}-4x{+4}+{y}^{2}-2y{+1}{-5}-4=0$

I added $5$ so I had to substract 5 to keep the equation balanced.

Now we can write the expressions with $x$ and $y$ as squares:

${(x-2)}^{2}+{(y-1)}^{2}-9=0$

Finally I can move $9$ to right side to get the form (1)

${(x-2)}^{2}+{(y-1)}^{2}=9$

From this equation I can read that the equation shows a circle with center in $C=(2;1)$ and radius $r=3$

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