Question

show that the equation represents a circle by rewriting it in standard form, and find the center and radius of the circle. x^2 + y^2 - 6x + 10y + 30 = 0 standard form center (x, y) = ( ) radius

Explanation:

Step1: Complete the square for x - terms

The given equation is $x^{2}+y^{2}-6x + 10y+30 = 0$. For the $x$ - terms $x^{2}-6x$, we know that $(x - a)^{2}=x^{2}-2ax + a^{2}$. Here, $-2a=-6$, so $a = 3$ and $x^{2}-6x=(x - 3)^{2}-9$.

Step2: Complete the square for y - terms

For the $y$ - terms $y^{2}+10y$, since $(y + b)^{2}=y^{2}+2by + b^{2}$ and $2b = 10$ (so $b = 5$), then $y^{2}+10y=(y + 5)^{2}-25$.

Step3: Rewrite the equation

Substitute the completed - square expressions into the original equation:
\[

$$\begin{align*} (x - 3)^{2}-9+(y + 5)^{2}-25+30&=0\\ (x - 3)^{2}+(y + 5)^{2}-9-25 + 30&=0\\ (x - 3)^{2}+(y + 5)^{2}-4&=0\\ (x - 3)^{2}+(y + 5)^{2}&=4 \end{align*}$$

\]
The standard form of a circle is $(x - h)^{2}+(y - k)^{2}=r^{2}$, where $(h,k)$ is the center of the circle and $r$ is the radius.

Answer:

Center: $(3,-5)$
Radius: $2$
Standard form: $(x - 3)^{2}+(y + 5)^{2}=4$