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The equation of a circle which has a center at $(-5,2)$ can be written as $Ax^2 + 2y^2 + Bx + Cy = 40.$ Let $r$ be the radius of the circle. Find $A+B+C+r.$

Creeperhissboom  Apr 20, 2018
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 #1
avatar+474 
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I'll try to solve it.

ant101  Apr 20, 2018
edited by ant101  Apr 20, 2018
 #2
avatar+92437 
+3

The equation of a circle which has a center at (-5,2) can be written as Ax^2 + 2y^2 + Bx + Cy = 40

 Let r be the radius of the circle.

Find     A+B+C+r

 

\((x+5)^2+(y-2)^2=r^2\\ x^2+10x+25+y^2-4y+4=r^2\\ x^2+y^2+10x-4y=r^2-29\\ 2x^2+2y^2+20x-8y=2r^2-58\\ A=2,\quad B=20,\quad C=-8\\ 2r^2-58=40\\ r^2-29=20\\ r^2=49\\ r=7 \qquad \text{Since the radius must be positive} \)

 

You can do the addition. 

Melody  Apr 20, 2018

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