Final answer:
The equation of the circle centered at (5,6) with radius 7 is x^2 + y^2 - 10x - 12y - 12 = 0. The line segment joining (6,-2) and (3,4) is not a diameter of the circle.
Step-by-step explanation:
To find the equation of a circle centered at (5,6) and having a radius of 7, we can use the standard form equation of a circle: (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center of the circle and r is the radius.
Substituting the given values, we have (x - 5)^2 + (y - 6)^2 = 7^2. Expanding this equation, we get x^2 - 10x + 25 + y^2 - 12y + 36 = 49. Simplifying further, the equation of the circle is x^2 + y^2 - 10x - 12y - 12 = 0.
To check if the line segment joining (6, -2) and (3, 4) is a diameter of the circle, we can find the distance between the two points. Using the distance formula, we get sqrt((3 - 6)^2 + (4 - (-2))^2) = sqrt((-3)^2 + 6^2) = sqrt(45) = 3sqrt(5).
Since the diameter of the circle is twice the radius, which is 2(7) = 14, and the distance between the two given points is 3sqrt(5) ≪ 14, the line segment is not a diameter of the circle.