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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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