+0  
 
0
284
1
avatar

Find the area of the region enclosed by the graph of $x^2 + y^2 = 2x - 6y + 6 - 18x + 2y$

 Oct 28, 2021
 #1
avatar+677 
0

Hello Guest,

 

\(x^2+y^2=2-x-6y+6-18x+2y\)

\(\mbox {Move the variables to the left side and change their signs}\)

\(x^2+y^2-2x+6y+18x-2y=6\)

\(x^2+y^2+16x+6y-2y=6\)

\(x^2+y^2+16x+4y=6\)

\(\mbox {Use the commutative property to reorder the terms}\)

\(x^2+16x+y^2+4y=6\)

\(\mbox {To complete the square, the same value needs to be added to both sides}\)

\(x^2+16x+?+y^2+4y=6+?\)

\(\mbox {To complete the square } x^2+16x+64=(x+8)^2 \mbox { add 64 to the expression}\)

\(x^2+16x+64+y^2+4y=6+?\)

\(\mbox {Since 64 was added to the left side, also add 64 to the right side}\)

\(x^2+16x+64+y^2+4y=6+64\)

\(\mbox {Use the first binomial } a^2+2ab+b^2=(a+b)^2 \mbox { to factor the expression}\)

\((x+8)^2+y^2+4y=6+64\)

\((x+8)^2+y^2+4y=70\)

\(\mbox {To complete the square, the same value needs to be added to both sides}\)

\((x+8)^2+y^2+4y+?=70+?\)

\(\mbox {To complete the square } y^2+4y+4=(y+2)^2 \mbox { add 4 to the expression}\)

\((x+8)+y^2+4y+4=70+?\)

\(\mbox {Since 4 was added to the left side, also add 4 to the right side}\)

\((x+8)+y^2+4y+4=70+4\)

\(\mbox {Use the first binomial } a^2+2ab+b^2=(a+b)^2 \mbox { to factor the expression}\)

\((x+8)^2+(y+2)^2=70+4\)

\((x+8)^2+(y+2)^2=74\)

\(\mbox {The equation can be written in the form } (x-p)^2+(y-q)^2=r^2 \mbox { , so it represents a circle with the radius } r=\sqrt{74} \mbox { and the center } (-8,-2)\)

 Oct 29, 2021

8 Online Users

avatar
avatar
avatar
avatar
avatar
avatar
avatar