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Find the value of C so that the graph of x2 + y2 - 12x + 14y - C = 0 
passes through the point (10,1)

 Feb 12, 2022
 #1
avatar+36434 
+1

x^2 -12x       + y^2 + 14y        = C

(x-6)^2          + (y+7)^2          =     C                       Circle centered at   6, -7 

                                                                                 distance fomula to 10,1 is the radius

                                                                                           4^2 + 8^2 = d^2 = r^2

                                                                                            80 = r^2         C needs to equal 80

(x-6)^2     + (y+7)^2          - 80   = 0   

 Feb 12, 2022
edited by Guest  Feb 12, 2022
edited by Guest  Feb 12, 2022
 #2
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+1

It says it's wrong.

Guest Feb 12, 2022
 #3
avatar+36434 
+1

Here is a graph

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

I think we need to get it in the correct form:

 

(x-6)^2     + (y+7)^2          - 80   = 0  

x^2 -12x + 36    + y^2 + 14y + 49    - 80 = 0         THEN  - C  = 36 + 49 - 80 =   + 5       so    C = -5     

ElectricPavlov  Feb 12, 2022
edited by ElectricPavlov  Feb 12, 2022
 #4
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+1

Thank You! I get it now.

Guest Feb 12, 2022

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