heureka

Find an equation of the circle that satisfies the given conditions.

$$\boxed{x^2+y^2+2mx+2ny+q=0}$$

Find the standard form:

$$\\x^2+2mx+y^2+2ny=-q\\ x^2+\textcolor[rgb]{1,0,0}{2m}x+y^2+\textcolor[rgb]{0,0,1}{2n}y=-q\\ \underbrace{x^2+\textcolor[rgb]{1,0,0}{2m}x+ (\frac{ \textcolor[rgb]{1,0,0}{2m} }{2})^2 }_{ (x+m)^2} + \underbrace{y^2+\textcolor[rgb]{0,0,1}{2n}y+( \frac{ \textcolor[rgb]{0,0,1}{2n} }{2})^2}_{ (y+n)^2} =-q +(\frac{ \textcolor[rgb]{1,0,0}{2m} }{2})^2 +(\frac{ \textcolor[rgb]{0,0,1}{2n} }{2})^2\\ (x+m)^2+(y+n)^2=m^2+n^2-q\\ \boxed{ \Rightarrow x_0 = -m \qquad y_0=-n \qquad R = \sqrt{m^2+n^2-q}}$$

Hi sally1,

$$\boxed{x^2+y^2-6x+6y+9=0}$$

$$compare: ax^2+2bxy+cy^2+2dx+2ey+f=0$$

$$b=0 \quad and \quad a=c \quad
\Rightarrow \textcolor[rgb]{1,0,0}{\mbox{this is a circle}}$$

$$\mbox{compare for a circle}: x^2+y^2+2mx+2ny+q=0$$

$$2m=-6 \quad and \quad 2n=6 \\
\Rightarrow \\
m=-3 \\
n=3\\
q=9$$

$$\\radius: R = \sqrt{m^2+n^2-q}=\sqrt{(-3)^2+3^2-9}=\sqrt{9}=3\\
Center: x_0=-m=3 \quad and \quad y_0=-n=-3$$

radius = 3

center (3, -3)

$$\boxed{x^2+y^2+2mx+2ny+q=0}$$

Find the standard form:

$$\\x^2+2mx+y^2+2ny=-q\\
x^2+\textcolor[rgb]{1,0,0}{2m}x+y^2+\textcolor[rgb]{0,0,1}{2n}y=-q\\
\underbrace{x^2+\textcolor[rgb]{1,0,0}{2m}x+
(\frac{ \textcolor[rgb]{1,0,0}{2m} }{2})^2 }_{
(x+m)^2}
+
\underbrace{y^2+\textcolor[rgb]{0,0,1}{2n}y+(
\frac{ \textcolor[rgb]{0,0,1}{2n} }{2})^2}_{
(y+n)^2} =-q
+(\frac{ \textcolor[rgb]{1,0,0}{2m} }{2})^2
+(\frac{ \textcolor[rgb]{0,0,1}{2n} }{2})^2\\
(x+m)^2+(y+n)^2=m^2+n^2-q\\
\boxed{
\Rightarrow x_0 = -m \qquad y_0=-n \qquad
R = \sqrt{m^2+n^2-q}}$$

$${\mathtt{789}}{\mathtt{\,-\,}}{\mathtt{296}} = {\mathtt{493}}$$

Sorry

I am not able well enough in English

heureka

Hi Alan,

thank you.

Please proof your equation.

Many Greetings

heureka

$$\mbox{Find an equation of the line that satisfies the given conditions.}\\
\mbox{Through (1/2, -2/7) perpendicular to the line 5x-10y=1}$$

$$\\5x-10y=1 \quad \Rightarrow \quad g(x) = 0.5x-0.1\\
f(x)=ax^2+bx+c\\
\boxed{\mbox{1. }f(x)=g(x)}\quad and \quad \boxed{\mbox{2. }f'(x)*g'(x)=-1}$$

$$f(x)=ax^2+bx+c\\
2=a*1^2+b*1+c\\
7=a*(-2)^2+b(-2)+c\\
(2)\quad 2=a+b+c\\
(1)\quad 7=4a-2b+c\\
(1)-(2)\quad 5=3a-3b \quad \Rightarrow \boxed{b=a-\frac{5}{3}\mbox{ and }c=\frac{11}{3}-2a}$$

$$f(x)=g(x)\\
ax^2+bx+c=0.5x-0.1\\
ax^2+(a-\frac{5}{3})x+\frac{11}{3}-2a=0.5x-0.1 \\
\Rightarrow \boxed{x_{1,2}=\frac{13-6a}{12a}\pm\sqrt{\frac{(6a-13)^2}{(12a)^2}-\frac{(113-60a)}{30a}}}$$

$$f'(x)*g'(x)=-1\\
f'(x)=2ax+b\\
g'(x)=0.5\\
(2ax+b)*0.5=-1 \quad b=a-\frac{5}{3}\\
\Rightarrow 2ax+b=-2\\
\Rightarrow 2ax+(a-\frac{5}{3})=-2\\
\boxed{2ax+a=-\frac{1}{3}}$$

$$2a\left[ \frac{13-6a}{12a}\pm\sqrt{\frac{(6a-13)^2}{(12a)^2}-\frac{113-60a}{30a}
}\right]+a=-\frac{1}{3}\\
\Rightarrow 324a^2-542.4a-56=0\\
\Rightarrow \boxed{a_{1,2}=\frac{22.6\pm\sqrt{636.76}}{27}
}$$

$$\mbox{a must be negative: }
\boxed{a=\frac{22.6-\sqrt{636.76}}{27}}$$

$${\mathtt{a}} = {\frac{\left({\mathtt{22.6}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{636.76}}}}\right)}{{\mathtt{27}}}} \Rightarrow {\mathtt{a}} = -{\mathtt{0.097\: \!559\: \!403\: \!970\: \!835\: \!4}}$$

$${\mathtt{b}} = {\frac{\left({\mathtt{22.6}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{636.76}}}}\right)}{{\mathtt{27}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{5}}}{{\mathtt{3}}}} \Rightarrow {\mathtt{b}} = -{\mathtt{1.764\: \!226\: \!070\: \!637\: \!502\: \!1}}$$

$${\mathtt{c}} = {\frac{{\mathtt{11}}}{{\mathtt{3}}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}\left({\frac{\left({\mathtt{22.6}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{636.76}}}}\right)}{{\mathtt{27}}}}\right) \Rightarrow {\mathtt{c}} = {\mathtt{3.861\: \!785\: \!474\: \!608\: \!337\: \!5}}$$

The eqaution is:

$$\\f(x)=ax^2+bx+c\\
\boxed{f(x)=-0.0975594039708354x^2-1.7642260706375021x+3.8617854746083375
}$$

3x+2y=-6

$$\boxed{y=0}$$

3x = -6 | :3

$$\boxed{x=-2}$$

x-intercept is -2

Matrix A:
( 1,  1,  1,  1,  1,

  0,  1,  3,  2,  0, 

  1,  1,  0,  5,  1,

  0,  0,  1, -4,  2)

Matix A: toReducedRowEchelonForm
( 1,  0,  0,  -9,  0,

  0,  1,  0, 14,  0,

  0,  0,  1,  -4,  0,

  0,  0,  0,   0,  1)