(a) To find the equation of the circle, we first determine the center. The perpendicular bisectors of the chords \(AB\) and \(BC\) intersect at the center of the circle. The midpoint of \(AB\) is \((7, 5)\) and the midpoint of \(BC\) is \((4, 9)\). The center of the circle is \((4, 5)\).
The radius \(r\) is the distance from the center \((4, 5)\) to any of the points on the circle, such as \(A(7, 1)\):
\(r^2 = (7-4)^2 + (1-5)^2 = 3^2 + (-4)^2 = 9 + 16 = 25\)
Thus, the radius \(r = 5\).
The equation of the circle is \((x-4)^2 + (y-5)^2 = 25\).
(b) The gradient of the radius at \(B\) is:
\(\text{Gradient of radius} = \frac{9-5}{7-4} = \frac{4}{3}\)
The tangent at \(B\) is perpendicular to the radius, so its gradient is the negative reciprocal:
\(\text{Gradient of tangent} = -\frac{3}{4}\)
The equation of the tangent line at \(B(7, 9)\) is:
\(y - 9 = -\frac{3}{4}(x - 7)\)
Simplifying gives:
\(y = -\frac{3}{4}x + \frac{57}{4}\)
