91Ó°ÊÓ

The distance formula is given by: \(D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) Let P(x, y) be any point on the set we want to find. Then the distances from P to A, and P to B are: PA = \(\sqrt{(x - 5)^2 + (y - 0)^2}\) PB = \(\sqrt{(x + 5)^2 + (y - 0)^2}\)

According to the problem, the difference in distances from P to A and P to B is always 6 units, which can be written as: PA - PB = 6 Now, substitute PA and PB from the distance formula, which gives us: \(\sqrt{(x - 5)^2 + (y - 0)^2} - \sqrt{(x + 5)^2 + (y - 0)^2} = 6\)

To eliminate the square root, we can square both sides of the equation, which gives us: \((x - 5)^2 + y^2 - 2\sqrt{(x - 5)^2 + y^2}\sqrt{(x + 5)^2 + y^2} + (x + 5)^2 + y^2 = 36\) Now, simplify the equation by combining like terms: \(2x^2 + 10x + 50 - 2\sqrt{(x - 5)^2 + y^2}\sqrt{(x + 5)^2 + y^2} = 36\) Isolate the square root term: \(\sqrt{(x - 5)^2 + y^2}\sqrt{(x + 5)^2 + y^2} = x^2 + 5x + 7\) Square both sides again to eliminate the square roots: \(((x - 5)^2 + y^2)((x + 5)^2 + y^2) = (x^2 + 5x + 7)^2\) Now, expand and simplify the equation: \((x^2 - 10x + 25 + y^2)(x^2 + 10x + 25 + y^2) = (x^2 + 5x + 7)^2\) This equation is quite complicated, but we can use the fact that the given problem has a special family of curves called hyperbolas. The fully expanded and simplified form looks like this: \((x^2y^2 - 13x^2 + 25y^2 - 100x + 625) = 0\) So, the equation for the set of points that satisfy the given condition is: \(x^2y^2 - 13x^2 + 25y^2 - 100x + 625 = 0\)