+0  
 
0
38
3
avatar

The equation of a circle which has a center at (-5, 2) can be written as Ax^2 + 2y^2 + Bx + Cy = \(80\). Find the radius of the circle

 Apr 15, 2022

Best Answer 

 #1
avatar+55 
-1

Since the center is at (-5, 2), The equation of the circle is \((x+5)^2+(y-2)^2=r^2\)

Then you can multiply out \(x^2+10x+25+y^2-4y+4=r^2\)

\(x^2+y^2+10x-4y+29=r^2 \\ 2x^2+2y^2+20x-8y=2r^2-58 \\ 2r^2-58=80 \\ r^2-29=40 \\ r^2=69 \\ r=\sqrt{69} \)

(r has to be positive)

 Apr 15, 2022
edited by qjin27  Apr 15, 2022
 #1
avatar+55 
-1
Best Answer

Since the center is at (-5, 2), The equation of the circle is \((x+5)^2+(y-2)^2=r^2\)

Then you can multiply out \(x^2+10x+25+y^2-4y+4=r^2\)

\(x^2+y^2+10x-4y+29=r^2 \\ 2x^2+2y^2+20x-8y=2r^2-58 \\ 2r^2-58=80 \\ r^2-29=40 \\ r^2=69 \\ r=\sqrt{69} \)

(r has to be positive)

qjin27 Apr 15, 2022
edited by qjin27  Apr 15, 2022
 #2
avatar+55 
-3

I'm not sure that this is correct so can anyone double check?

qjin27  Apr 15, 2022
 #3
avatar+122390 
0

Looks good, qjin  !!!

 

cool cool cool

 Apr 15, 2022
edited by CPhill  Apr 15, 2022

10 Online Users

avatar