Answer:
Explanation:
1. Let x be the length of AD, which is also the length of AB because triangle ABD is isosceles. Since triangle ABC is a right triangle with angle BAC measuring 90 degrees, we have:
cos(BAC) = AB/AC
cos(90) = AB/8
0 = AB/8
AB = 0
2.This is not a possible length for AB, so the problem is not well-defined. There is no circle that passes through A, B, and C with center on the line AD.
Let O be the center of the circle. Since triangle OAB is isosceles, we have:
OA = OB
Let x be the length of OC. Then, by the Pythagorean theorem in triangle OAC, we have:
OA^2 + AC^2 = OC^2
Substituting OA = OB and AC = 6, we get:
OB^2 + 6^2 = x^2
Since triangle OBD is a right triangle, we also have:
OD^2 + DB^2 = OB^2
Substituting OD = OC and DB = 5, we get:
OC^2 + 5^2 = OB^2
We can solve this system of equations for OB^2:
OB^2 + 6^2 = x^2
OC^2 + 5^2 = OB^2
Adding the two equations, we get:
OC^2 + 6^2 + 5^2 = x^2
Substituting OC = x - 6, we get:
(x - 6)^2 + 6^2 + 5^2 = x^2
Simplifying and solving for x, we get:
x = 13
3.Therefore, the length of OC is 13. The radius of the circle is equal to the length of OA or OB, which is:
r = OB = sqrt(OC^2 - AB^2) = sqrt(13^2 - 6^2) = 5sqrt(2)
So the equation of the circle is:
(x - h)^2 + (y - k)^2 = r^2
Substituting (h, k) = (0, 0) and r = 5sqrt(2), we get:
x^2 + y^2 = 50
Therefore, the equation of the circle is x^2 + y^2 = 50.
Let O be the center of the circle. Since triangle AOB is isosceles, we have:
OA = OB
Let x be the length of OC. Then, by the Pythagorean theorem in triangle OAC, we have:
OA^2 + AC^2 = OC^2
Substituting OA = OB and AC = 6, we get:
OB^2 + 6^2 = x^2
Since triangle OCD is a right triangle, we also have:
OD^2 + DC^2 = OC^2
Substituting OD = OA and DC = 8, we get:
OA^2 + 8^2 = x^2
We can solve this system of equations for x:
OB^2 + 6^2 = x^2
OA^2 + 8^2 = x^2
Adding the two equations, we get:
2x^2 = 100
x^2 = 50
x = sqrt(50) = 5sqrt(2)
Therefore, the length of OC is 5sqrt(2). The radius of the circle is equal to the length of OA or OB, which is:
r = OB = sqrt(OC^2 - AB^2) = sqrt(50 - 36) = 2sqrt(14)
So the equation of the circle is:
(x - h)^2 + (y - k)^2 = r^2
Substituting (h, k) = (0, 0) and r = 2sqrt(14), we get:
x^2 + y^2 = 56
Therefore, the equation of the circle is x^2 + y^2 = 56.
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