Question

write the equation in standard form for the circle $x^{2}+y^{2}+10y - 3 = 0$.

Explanation:

Step1: Recall circle standard form

The standard form of a circle's equation is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius. We need to complete the square for the given equation \(x^2 + y^2 + 10y - 3 = 0\).

Step2: Group x and y terms

Group the \(x\)-terms and \(y\)-terms: \(x^2 + (y^2 + 10y) - 3 = 0\). For the \(x\)-terms, \(x^2\) is already a perfect square (since the coefficient of \(x\) is 0, so \((x - 0)^2\)).

Step3: Complete the square for y

For the \(y\)-terms \(y^2 + 10y\), we use the formula \((a + b)^2 = a^2 + 2ab + b^2\). Here, \(a = y\), \(2ab = 10y\), so \(2b = 10\) which means \(b = 5\). Then \(b^2 = 25\). We add and subtract 25 (but since we add 25 to complete the square, we need to add it to the other side to keep the equation balanced).

So, \(x^2 + (y^2 + 10y + 25 - 25) - 3 = 0\).

Step4: Rewrite the equation

Rewrite the \(y\)-terms as a perfect square and simplify the constants: \(x^2 + (y + 5)^2 - 25 - 3 = 0\).

Simplify the constants: \(x^2 + (y + 5)^2 - 28 = 0\).

Then, move the constant to the other side: \(x^2 + (y + 5)^2 = 28\).

We can also write the \(x\)-term as \((x - 0)^2\), so the equation is \((x - 0)^2 + (y + 5)^2 = 28\).

Answer:

\(\boldsymbol{(x - 0)^2 + (y + 5)^2 = 28}\) (or equivalently \(x^2 + (y + 5)^2 = 28\))