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The equation of the circle shown in the following diagram can be written as $x^2 + Ay^2 + Bx + Cy + D = 0$. Find $A+B+C+D$.

Guest Feb 22, 2018

Best Answer 

 #1
avatar+7056 
+2

the center of the circle =  (-1, 1)

 

the radius of the circle  =  \(\sqrt{(-1-1)^2+(1-2)^2}\,=\,\sqrt{(-2)^2+(-1)^2}\,=\,\sqrt{4+1}\,=\,\sqrt5\)

 

So the equation of the circle is...

 

\((x--1)^2+(y-1)^2\,=\,\sqrt5^2 \\~\\ (x+1)^2+(y-1)^2\,=\,5 \\~\\ (x+1)(x+1)+(y-1)(y-1)\,=\,5 \\~\\ x^2+2x+1+y^2-2y+1\,=\,5 \\~\\ x^2+2x+y^2-2y+2\,=\,5 \\~\\ x^2+2x+y^2-2y-3\,=\,0 \\~\\ x^2+y^2+2x-2y-3\,=\,0 \)


A + B + C + D   =   1 + 2 - 2 - 3   =   -2

 

Here's a graph to check the equation of the circle:

https://www.desmos.com/calculator/rnauqvgqrg

hectictar  Feb 22, 2018
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1+0 Answers

 #1
avatar+7056 
+2
Best Answer

the center of the circle =  (-1, 1)

 

the radius of the circle  =  \(\sqrt{(-1-1)^2+(1-2)^2}\,=\,\sqrt{(-2)^2+(-1)^2}\,=\,\sqrt{4+1}\,=\,\sqrt5\)

 

So the equation of the circle is...

 

\((x--1)^2+(y-1)^2\,=\,\sqrt5^2 \\~\\ (x+1)^2+(y-1)^2\,=\,5 \\~\\ (x+1)(x+1)+(y-1)(y-1)\,=\,5 \\~\\ x^2+2x+1+y^2-2y+1\,=\,5 \\~\\ x^2+2x+y^2-2y+2\,=\,5 \\~\\ x^2+2x+y^2-2y-3\,=\,0 \\~\\ x^2+y^2+2x-2y-3\,=\,0 \)


A + B + C + D   =   1 + 2 - 2 - 3   =   -2

 

Here's a graph to check the equation of the circle:

https://www.desmos.com/calculator/rnauqvgqrg

hectictar  Feb 22, 2018

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