Find the value of C so that the graph of x2 + y2 - 12x + 14y - C = 0
passes through the point (10,1)
+37157 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
+37157 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
