(a) Given three points \(A(7,1)\), \(B(7,9)\), and \(C(1,9)\) on the circumference of the circle, we can use the fact that the center of the circle is equidistant from these points.
The center of the circle \((h, k)\) is the midpoint of the line segment connecting any two points on the circumference. Let’s use \(A\) and \(C\) to find the center:
\( h = \frac{7 + 1}{2} = 4 )
\(  k = \frac{1 + 9}{2} = 5 \)
So, the center of the circle is \((4, 5)\).
Now, we need to find the radius (\(r\)), which is the distance from the center to any of the given points (let’s use \(A\)):
\( r = \sqrt{(7 – 4)^2 + (1 – 5)^2} = \sqrt{(3)^{2}+(-4)^2} = 5 \)
\( (x – 4)^2 + (y – 5)^2 = 5^2 \)
\( (x – 4)^2 + (y – 5)^2 = 25 \)
(b) The gradient of the radius from the center of the circle to point \(B(7, 9)\) is calculated as:
\( \text{Gradient of radius} = \frac{y_B – y_C}{x_B – x_C} = \frac{9 – 5}{7 – 4} = \frac{4}{3} \)
Now, for the equation of the tangent, we can use the fact that the radius and the tangent at the point of contact are perpendicular. Therefore, the negative reciprocal of the gradient of the radius is the gradient of the tangent.
\( \text{Gradient of tangent} = -\frac{1}{\text{Gradient of radius}} = -\frac{1}{\frac{4}{3}} = -\frac{3}{4} \)
Now, using the point-slope form of a line (\(y – y_1 = m(x – x_1)\)),
where \((x_1, y_1)\) is a point on the line and \(m\) is the gradient,
The equation of the tangent at \(B(7, 9)\):
\( y – 9 = -\frac{3}{4}(x – 7) \)
\( y – 9 = -\frac{3}{4}(x – 7) \)