Question

1. find the center and radius of the circle with the given equation 4x² + 4y² + 24x - 16y + 40 = 0

Explanation:

Step1: Divide the equation by 4

Divide the entire equation $4x^{2}+4y^{2}+24x - 16y+40 = 0$ by 4 to simplify. We get $x^{2}+y^{2}+6x - 4y + 10=0$.

Step2: Complete the square for x - terms

For the $x$ - terms in $x^{2}+6x$, we know that $(x + a)^{2}=x^{2}+2ax+a^{2}$. Here $2a = 6$, so $a = 3$ and $x^{2}+6x=(x + 3)^{2}-9$.

Step3: Complete the square for y - terms

For the $y$ - terms in $y^{2}-4y$, we have $2a=-4$, so $a=-2$ and $y^{2}-4y=(y - 2)^{2}-4$.

Step4: Rewrite the equation

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

$$\begin{align*} (x + 3)^{2}-9+(y - 2)^{2}-4+10&=0\\ (x + 3)^{2}+(y - 2)^{2}-9-4 + 10&=0\\ (x + 3)^{2}+(y - 2)^{2}-3&=0\\ (x + 3)^{2}+(y - 2)^{2}&=3 \end{align*}$$

\]
The standard form of a circle equation 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: $(-3,2)$; Radius: $\sqrt{3}$