Question

find the angle between x²-y²-12x-6y+41=0, x²+y²+4x+6y-56​

Answer 1

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Answer 2

x^2 - 6x +y^2 - 4y = 12

Then take 1/2 of the 'b' term for both quadratic expressions, square those values and add them to both sides.

#x^2 -6x + 9 + y^2 - 4y + 4 = 12 + 9 + 4

(x - 3)^2 + (y -2)^2 = 25

Circle centered at (3,2) with radius = 5