Let a right-angled triangle ABC with A = 2 be inscribed in the circle x^2 + y^2 - 8x - 10y + k = 0 . If the area of the ABC is 24 and the length of the side AB is 6 , then the value of the constant k is equal to:

Step-by-step solution

Given the area of the right-angled ABC is 24 and AB = 6 . The area of the triangle is given by 1 2 AB AC = 24 . Substituting the value of AB , we get 1 2 6 AC = 24 AC = 8 . Since A = 2 , the triangle is right-angled at A , which means the hypotenuse BC acts as the diameter of the circumscribing circle. Using Pythagoras theorem, BC = AB^2 + AC^2 = 6^2 + 8^2 = 36 + 64 = 10 . The radius of the circle r is half of the diameter, so r = 10 2 = 5 . The given equation of the circle is x^2 + y^2 - 8x - 10y + k = 0 . The rad