(a) Substitute the line equation \(y = 3x - 20\) into the circle equation \((x+1)^2 + (y-2)^2 = 85\).

\((x+1)^2 + (3x-22)^2 = 85\)

Expand and simplify: \(10x^2 - 130x + 400 = 0\)

Factorize: \((x-8)(x-5) = 0\)

Solutions: \(x = 8\) or \(x = 5\)

For \(x = 8\), \(y = 3(8) - 20 = 4\), so \((8, 4)\)

For \(x = 5\), \(y = 3(5) - 20 = -5\), so \((5, -5)\)

(b) Mid-point of \(AB\) is \(\left(\frac{6.5}{2}, -\frac{1}{2}\right)\)

Use \(C = (-1, 2)\)

Calculate \(r^2\): \((-1 - \frac{6.5}{2})^2 + (2 + \frac{1}{2})^2 = 62\frac{1}{2}\)

Equation of the circle: \((x+1)^2 + (y-2)^2 = 62\frac{1}{2}\)

(a) Substitute \(y = 2x + 5\) into \(x^2 + y^2 = 20\) to get \(x^2 + (2x + 5)^2 = 20\).

Expand to get \(x^2 + 4x^2 + 20x + 25 = 20\), leading to \(5x^2 + 20x + 25 = 20\).

Simplify to \(5x^2 + 20x + 5 = 0\) and divide by 5: \(x^2 + 4x + 1 = 0\).

Use the quadratic formula: \(x = \frac{-4 \pm \sqrt{16 - 4}}{2}\).

Calculate \(x = -2 \pm \sqrt{3}\).

Substitute back to find \(y\): \(y = 2(-2 + \sqrt{3}) + 5 = 1 + 2\sqrt{3}\) and \(y = 2(-2 - \sqrt{3}) + 5 = 1 - 2\sqrt{3}\).

Coordinates are \(A = (-2 + \sqrt{3}, 1 + 2\sqrt{3})\) and \(B = (-2 - \sqrt{3}, 1 - 2\sqrt{3})\).

Use the distance formula: \(AB = \sqrt{((-2 + \sqrt{3}) - (-2 - \sqrt{3}))^2 + ((1 + 2\sqrt{3}) - (1 - 2\sqrt{3}))^2}\).

Simplify to \(AB = \sqrt{(2\sqrt{3})^2 + (4\sqrt{3})^2} = \sqrt{12 + 48} = \sqrt{60}\).

(b) The equation of the tangent is \(y = m(x - 10)\).

Substitute into the circle equation: \(x^2 + (m(x - 10))^2 = 20\).

Expand to get \(x^2 + m^2(x^2 - 20x + 100) = 20\).

Collect terms: \((m^2 + 1)x^2 - 20m^2x + 100m^2 = 20\).

Use the discriminant condition for tangency: \(b^2 - 4ac = 0\).

Calculate \(400m^4 - 80(5m^2 + 4m^2 - 1) = 0\).

Simplify to \(-80(4m^2 - 1) = 0\), leading to \(m = \pm \frac{1}{2}\).

(a) The center of the circle is found by completing the square: \((x-3)^2 + (y+2)^2 = 40\), so the center is \((3, -2)\) and the radius is \(\sqrt{40}\).

The gradient of the radius at \(P (5, 4)\) is \(\frac{-2 - 4}{3 - 5} = 3\). The gradient of the tangent is the negative reciprocal, \(-\frac{1}{3}\).

The equation of the tangent line is \(y - 4 = -\frac{1}{3}(x - 5)\), which simplifies to \(y = -\frac{1}{3}x + \frac{17}{3}\).

Intercepts: \(A (17, 0)\) and \(B (0, \frac{17}{3})\).

Area of \(\triangle OAB = \frac{1}{2} \times 17 \times \frac{17}{3} = \frac{289}{6}\).

(b) The radius of the circle is \(\sqrt{40}\).

For \(\triangle PQR\) to be equilateral, each side is \(\sqrt{40}\). The area of \(\triangle PQR\) is \(\frac{\sqrt{3}}{4} \times (\sqrt{40})^2 = 30\sqrt{3}\).

(a) The equation of a circle with center (h, k) and radius r is given by \((x - h)^2 + (y - k)^2 = r^2\). Here, the center is (5, 2) and the circle passes through (7, 5).

Calculate the radius: \(r^2 = (7 - 5)^2 + (5 - 2)^2 = 2^2 + 3^2 = 4 + 9 = 13\).

Thus, the equation of the circle is \((x - 5)^2 + (y - 2)^2 = 13\).

(b) Substitute \(y = 5x - 10\) into the circle's equation: \((x - 5)^2 + ((5x - 10) - 2)^2 = 13\).

Simplify: \((x - 5)^2 + (5x - 12)^2 = 13\).

Expand and simplify: \(26x^2 - 130x + 156 = 0\).

Factorize: \(26(x - 2)(x - 3) = 0\).

Solutions are \(x = 2\) and \(x = 3\).

Find corresponding y-values: \(y = 5(2) - 10 = 0\) and \(y = 5(3) - 10 = 5\).

Points of intersection are (2, 0) and (3, 5).

Length of chord AB: \(AB = \sqrt{(3 - 2)^2 + (5 - 0)^2} = \sqrt{1 + 25} = \sqrt{26}\).

(a) Calculate the gradients of \(AB\) and \(BC\):

Gradient of \(AB = \frac{-3}{5}\), Gradient of \(BC = \frac{5}{3}\).

The product of the gradients \(m_{AB} \times m_{BC} = -1\), confirming \(\angle ABC = 90^\circ\).

(b) The center \(D\) is the midpoint of \(AC\):

\(D = \left( \frac{-2+6}{2}, \frac{3+5}{2} \right) = (2, 4)\).

(c) Use the circle equation \((x - x_c)^2 + (y - y_c)^2 = r^2\):

\((x-2)^2 + (y-4)^2 = 17\).

(d) Find \(E\) such that \(BE\) is a diameter:

\(E = (1, 8)\).

The gradient of the tangent is the negative reciprocal of \(BE\):

Gradient of \(BE = -4\), so gradient of tangent = \(\frac{1}{4}\).

Equation of the tangent: \(y - 8 = \frac{1}{4}(x - 1)\).