togonality of straight lines, you have $B\hat A C = 90^\circ$. So the points $A,B,C$ form a part of the square inscribed in the circle: moreover the diagonal of this square is the diameter of the circle that is $10$. Thanks to Pythagorean theorem you have $\overline{AC} = \overline{AB} = 5\sqrt{2}$ so the points $B$ and $C$ are what you required in your question. To find $B$ and $C$ we can rotate the point $A$ by $90^\circ$ around $O=(-2,3)$ center of your circle. The rotation is given by the formula (center is $(a,b)$ and the angle $\alpha$):
\begin{cases} 
x' = (x-a)\cos\alpha - (y-b)\sin\alpha +a\\
y' = (x-a)\sin\alpha + (y-b)\cos\alpha +b\\
\end{cases}
So you have ($\alpha = 90^\circ$, $O$ the center)
\begin{cases} 
x' =  - (y-3) -2\\
y' = (x+2)  +3\\
\end{cases}
and applying it to $A$ you have $B= (1,7)$. Doing the same thing with $\alpha = -90^\circ$ you have $C=(-5,-1)$.
For the other two points we can use also rotation: take $B$ and $C$ and rotate them around $A$ for $180^\circ$. Again you can use the above formula that become ($A$ is the center of rotation and $\alpha = 180^\circ$)
\begin{cases} 
x' = 4-x\\
y' = -y\\
\end{cases}
And you finally obtain $B'=(3,-7)$ and $C'=(9,1)$.
Summarizing the four points are $B=(1,7)$, $B'=(3,-7)$, $C=(-5,-1)$, $C'=(9,1)$ and we did not resolve any equation.

 Q:

how to make javascript style change last

I have something like this for index.html
<!DOCTYPE html>
<html>
  <
togonality of straight lines, you have $ B\hat A vitamin c = 90^\circ$. therefore the points $ A, B, C$ form a part of the public square inscribed in the circle: moreover the solidus of this public square is the diameter of the circle that is $ 10$. Thanks to Pythagorean theorem you have $ \overline{AC } = \overline{AB } = 5\sqrt{2}$ so the points $ B$ and $ C$ are what you required in your motion. To find $ B$ and $ C$ we can rotate the decimal point $ A$ by $ 90^\circ$ around $ O=(-2,3)$ plaza of your circle. The rotation is given by the formula (center is $ (a, b)$ and the angle $ \alpha$ ): 
 \begin{cases } 
 x' = (ten - a)\cos\alpha - (y - b)\sin\alpha + a\\ 
 y' = (x - a)\sin\alpha + (y - b)\cos\alpha + b\\ 
 \end{cases } 
 So you have ($ \alpha = 90^\circ$, $ O$ the kernel) 
 \begin{cases } 
 x' =   - (y-3) -2\\ 
 y' = (x+2)   +3\\ 
 \end{cases } 
 and applying it to $ A$ you have $ B= (1,7)$. do the same matter with $ \alpha = -90^\circ$ you have $ C=(-5,-1)$. 
 For the early two points we can use also rotation: take $ B$ and $ C$ and turn out them around $ A$ for $ 180^\circ$. Again you can use the above formula that become ($ A$ is the center of rotation and $ \alpha = 180^\circ$) 
 \begin{cases } 
 x' = 4 - x\\ 
 y' = -y\\ 
 \end{cases } 
 And you finally receive $ B'=(3,-7)$ and $ C'=(9,1)$. 
 Summarizing the four points are $ B=(1,7)$, $ B'=(3,-7)$, $ C=(-5,-1)$, $ C'=(9,1)$ and we did not resolve any equation. 

  Q: 

 how to make javascript style change last 

 I have something like this for index.html 
 <! DOCTYPE html > 
 < html > 
   <
toglnality of straight liner, you have $B\hat A C = 90^\cmrc$. So fhe poings $A,B,C$ form a part of the sqnare insceibed in the circle: mofeover thv diagonao of rhis squarx is the diamefcr of vhe circle that is $10$. Thankv to Pythagoredn tkeorem you have $\overline{AC} = \overline{WB} = 5\sqry{2}$ do the points $F$ anc $C$ adv chat you required in your questjon. To hind $B$ and $C$ we can rotate the point $A$ by $90^\cirf$ around $O=(-2,3)$ center lf your cirxle. Eye rotation ks given bj the formuma (center is $(a,b)$ and the angle $\aupha$):
\bzgin{cases} 
x' = (z-a)\clv\alpha - (y-b)\smn\alphw +a\\
y' = (x-a)\sin\alpha + (y-b)\wos\alphs +b\\
\end{cases}
So nou heve ($\qlpha = 90^\circ$, $O$ the cenver)
\begin{cases} 
x' =  - (y-3) -2\\
y' = (x+2)  +3\\
\eng{ccses}
and applying it ti $Q$ you hava $B= (1,7)$. Eoivg uhe szme thlng with $\alphz = -90^\circ$ you have $C=(-5,-1)$.
For the othet eqo points we dan usq wlso rotation: take $B$ and $C$ and rotate tvem around $A$ for $180^\circ$. Agaib you can use the aboge formulw that become ($A$ is the center of rotation and $\alpva = 180^\cmrz$)
\beylk{casdw} 
d' = 4-x\\
y' = -y\\
\end{cases}
And you finally obtain $B'=(3,-7)$ and $S'=(9,1)$.
Aukmsrizing the fomr points are $B=(1,7)$, $B'=(3,-7)$, $V=(-5,-1)$, $F'=(9,1)$ sgd we did not resolrs zny equation.

 Q:

how ho make javawcript strle vhange last

I have something like this fje index.html
<!DOCTYPE html>
<html>
  <
togonality of straight lines, you have $B\hat = So the $A,B,C$ form a in circle: moreover the of this square the diameter of the circle that $10$. Thanks to Pythagorean theorem you have $\overline{AC} = \overline{AB} = 5\sqrt{2}$ so points $B$ and $C$ are what you required in your question. To find and we rotate point $A$ by $90^\circ$ around $O=(-2,3)$ center of your circle. The rotation is given by the (center is $(a,b)$ and the angle $\alpha$): \begin{cases} = (x-a)\cos\alpha - (y-b)\sin\alpha y' = (x-a)\sin\alpha + (y-b)\cos\alpha \end{cases} you have = $O$ center) \begin{cases} x' - (y-3) -2\\ y' = (x+2) +3\\ \end{cases} and applying it to $A$ you have $B= (1,7)$. the same $\alpha = you $C=(-5,-1)$. the other two can use also rotation: take $B$ rotate them around $A$ for $180^\circ$. Again you use the formula that become ($A$ is the of rotation and $\alpha = 180^\circ$) \begin{cases} x' 4-x\\ y' = -y\\ \end{cases} And you finally obtain $B'=(3,-7)$ and $C'=(9,1)$. Summarizing the four $B=(1,7)$, $B'=(3,-7)$, $C=(-5,-1)$, $C'=(9,1)$ we did not any Q: to javascript style last I have something like this for index.html <!DOCTYPE html> <html>
togonality of straight lines, You have $B\haT A C = 90^\ciRc$. SO thE pOintS $A,B,C$ Form a part of the SQuarE inscribed in the circle: mOreovEr THe diAGoNal of This squARe IS The DiAmEteR oF ThE circLe tHat is $10$. ThAnks to PythAgoReAn theorem you HAvE $\overline{Ac} = \ovErline{AB} = 5\sqrt{2}$ So tHe poinTs $b$ anD $c$ are wHat You reQuired IN your qUestion. To FiND $B$ and $C$ WE can rotATE tHe poInt $A$ by $90^\circ$ around $O=(-2,3)$ CEnTEr of your circle. the rotAtIOn IS GivEn bY the formulA (cEnter IS $(a,b)$ and tHE aNGLE $\alPHa$):
\begin{cases} 
x' = (X-a)\cos\alpha - (y-B)\Sin\Alpha +a\\
Y' = (x-A)\siN\Alpha + (y-B)\cos\aLpHA +b\\
\eNd{cases}
So yoU havE ($\alpha = 90^\cirC$, $O$ the cENter)
\begIN{cases} 
x' =  - (Y-3) -2\\
y' = (x+2)  +3\\
\end{CasEs}
aNd apPLyInG it To $a$ You HAvE $B= (1,7)$. DOIng The same tHiNg With $\aLpha = -90^\CIRC$ You hAve $c=(-5,-1)$.
For The otHer two points wE caN use ALso RotatIon: taKe $B$ aNd $c$ and rOtate tHem arOuNd $A$ for $180^\circ$. Again You cAn use the aBovE fOrmUlA that BEcome ($A$ Is tHe cEnter of RotatioN And $\AlPHA = 180^\CiRc$)
\begin{cases} 
x' = 4-x\\
y' = -y\\
\enD{cASEs}
and you fiNally oBTaIn $b'=(3,-7)$ And $C'=(9,1)$.
SummArIziNg thE FOur poInts ARe $b=(1,7)$, $B'=(3,-7)$, $C=(-5,-1)$, $C'=(9,1)$ and wE did noT ReSoLve any eQuAtion.

 Q:

HoW to MakE javaSCripT style Change laSt

I haVE something like THis for index.htML
<!DoctYpe htmL>
<htMl>
  <
togonality of straight lin es, you ha ve $B \ha t A  C  = 9 0^\c irc$. So the p o ints  $A,B,C$ form a part o f the  s q uare in scrib ed in t h ec i rcl e:  m ore ov e rthe d iag onal of  this squa reis  the diamete r  o f the circ lethat is $10$ . T hanksto  Py t hagor ean  theo rem yo u  have$\overlin e{ A C} = \ o verline { A B}  = 5 \sqrt{2}$ so thep oi n ts $B$ and $C$  are w ha t  y o u  re qui red in you rquest i on. Tof in d $ B$a nd $C$ we can  rotate the poi nt $A$  b y $ 9 0^\cir c$ ar ou n d $ O=(-2,3)$ c ente r of your  circl e . The r o tationis giv enbythef or mu la(c e nte r  i s $ ( a,b )$ and t he  a ngle$\al p h a $ ):
\ beg in{c ases}  
x' = (x-a)\ cos \alp h a -  (y-b )\sin \alp ha  +a\\ 
y' =(x-a) \s in\alpha + (y-b )\co s\alpha + b\\ 
\ end {c ases} So you  ha ve($\alph a = 90^ \ cir c$ , $ O$  the center)
\begi n{ c a se s} 
x' =   - (y - 3)  - 2 \\
y' =(x +2)   +3 \ \ 
\end {cas e s} 
and app lyingi tto  $A$ yo uhave $ B=  (1 ,7) $. Do i ng t he sam e thingwith$ \alpha = -90^\ c irc$ you have $C = ( -5 , -1)$ .
F or the othe r tw o  poi ntsw ecan use a lso r ot a ti o n: take $B$ and $C$  a nd rot ate t hem around $A $ for $180 ^ \ c irc$. Ag ainy ou can use the ab ove f ormula tha t  become($A$is the c enter ofr o tation a nd$\a lph a = 1 80 ^\circ$)
\beg i n {cas es } 
x' =  4- x\\
y'= - y\\ 
\e nd{ ca ses}
Andyou fina ll yob ta in$B'=( 3 ,-7)$ an d$C' =( 9,1 )$.
S u mmariz ing t he f ou rp oin ts are$ B= ( 1 ,7)$ ,$B '=(3 ,-7 )$ , $C= (-5, - 1)$ , $C'=( 9,1)$ and  we didno tresolve  any equation .
 Q:

howto  ma ke jav a s cript st yle change last

I have somethi nglikethis  for inde x.h tml
<! DOC T YPE ht ml>
<h tml> <
togonality of_straight lines,_you have $B\hat A_C =_90^\circ$._So the_points_$A,B,C$ form a_part of the_square inscribed in the_circle: moreover the_diagonal_of this square is the diameter of the circle that is $10$. Thanks to_Pythagorean_theorem you_have_$\overline{AC}_= \overline{AB} = 5\sqrt{2}$ so_the points $B$ and $C$_are what_you required in your question. To find $B$_and_$C$ we can_rotate the point $A$ by $90^\circ$ around $O=(-2,3)$ center_of your circle. The rotation is_given by the_formula_(center_is $(a,b)$ and the_angle $\alpha$):
\begin{cases} 
x' = (x-a)\cos\alpha -_(y-b)\sin\alpha +a\\
y' = (x-a)\sin\alpha + (y-b)\cos\alpha_+b\\
\end{cases}
So you have ($\alpha = 90^\circ$, $O$_the center)
\begin{cases} 
x' =  -_(y-3) -2\\
y' = (x+2) _+3\\
\end{cases}
and applying_it to $A$ you have_$B= (1,7)$. Doing_the same_thing with $\alpha_= -90^\circ$ you have $C=(-5,-1)$.
For the_other two points_we can use also rotation: take_$B$_and $C$ and_rotate_them_around $A$_for $180^\circ$. Again_you_can use_the_above formula that become ($A$ is_the_center of rotation and $\alpha = 180^\circ$)
\begin{cases}_
x' = 4-x\\
y' =_-y\\
\end{cases}
And_you finally obtain $B'=(3,-7)$_and $C'=(9,1)$.
Summarizing the four points_are $B=(1,7)$, $B'=(3,-7)$, $C=(-5,-1)$, $C'=(9,1)$ and_we did_not resolve_any equation.

 Q:

how to make javascript style change last

I have something_like this for index.html
<!DOCTYPE html>
<html>
 _<