\(O\) is \((0, 0)\) and \(P\) is \((6, -8)\), so the gradient is the change in \(y\) divided by the change in \(x\):

\(\frac{ -8 }{ 6 } = -\frac{4}{3}\)

A tangent meets the radius drawn to the point of contact at right angles, so the tangent at \(P\) is perpendicular to \(OP\).

Perpendicular gradients multiply to \(-1\), so turn the gradient of \(OP\) upside down and change its sign:

\(-1 \div (-\frac{4}{3}) = \frac{3}{4}\)

Work the gradient out properly and compare.

The gradient of the radius \(OP\) is \(\frac{ -15 }{ -20 } = \frac{3}{4}\)

The tangent meets the radius at right angles, so its gradient is

\(-1 \div \frac{3}{4} = -\frac{4}{3}\)

The tangent's gradient is \(-\frac{4}{3}\). That is exactly what the student wrote.

Answer: Yes

The tangent meets the radius \(OP\) at right angles, so start with the gradient of \(OP\), where \(O\) is the origin.

\(\frac{ 8 }{ -15 } = -\frac{8}{15}\)

Perpendicular gradients multiply to \(-1\), so the tangent has gradient

\(-1 \div (-\frac{8}{15}) = \frac{15}{8}\)

The tangent is \(y = \frac{15}{8}x + c\), and it passes through \((-15, 8)\), so those coordinates fit it:

\(8 = \frac{15}{8} \times (-15) + c\)

\(8 = -\frac{225}{8} + c\)

Add \(\frac{225}{8}\) to both sides:

\(c = \frac{289}{8}\)

The tangent is \(y = \frac{15}{8}x + \frac{289}{8}\)

Every point on the \(y\)-axis has \(x = 0\), so put \(x = 0\) into the equation of the tangent. The \(x\) term disappears and what is left is the number on the end.

The tangent crosses the \(y\)-axis at \((0, -\frac{125}{4})\).

Every point on the \(x\)-axis has \(y = 0\), so put \(y = 0\) into the equation of the tangent and solve for \(x\).

\(0 = -\frac{3}{4}x - \frac{125}{4}\)

\(-\frac{3}{4}x = \frac{125}{4}\)

\(x = \frac{125}{4} \div (-\frac{3}{4}) = -\frac{125}{3}\)

The tangent crosses the \(x\)-axis at \((-\frac{125}{3}, 0)\).

\(PQ\) is a diameter, so \(O\) is the midpoint of \(PQ\) and \(Q\) is the same distance from the origin as \(P\) but the other way.

Change the sign of both coordinates: \(Q\) is \((-12, 16)\).

Check it is on the circle: \((-12)^2 + (16)^2 = 400\)

\(OQ\) lies along the same straight line as \(OP\), so the two radii have the same gradient, and the two tangents — each perpendicular to its own radius — are parallel to each other.

The tangent at \(Q\) therefore has the same gradient as the tangent at \(P\), namely \(\frac{3}{4}\).

Now put \(Q(-12, 16)\) into \(y = \frac{3}{4}x + c\):

\(16 = \frac{3}{4} \times (-12) + c\)

\(16 = -9 + c\)

Add \(9\) to both sides:

\(c = 25\)

The tangent at \(Q\) is \(y = \frac{3}{4}x + 25\)

The line \(PR\) has equation \(y = -\frac{1}{2}x + c\) and passes through \(R(-12, 10)\), so

\(10 = (-\frac{1}{2}) \times (-12) + c\)

\(c = 4\)

\(P\) is on the \(y\)-axis, so \(P = (0, 4)\). \(PQ\) is horizontal, so \(Q\) also has \(y = 4\).

Angle \(PRQ = 90^\circ\), so \(RQ\) is perpendicular to \(PR\), and its gradient is

\(-1 \div (-\frac{1}{2}) = 2\)

The line \(RQ\) passes through \(R(-12, 10)\):

\(y - 10 = 2(x + 12)\)

At \(Q\), \(y = 4\):

\(-6 = 2(x + 12)\)

\(x + 12 = -6 \div (2) = -3\)

\(x = -3 - 12 = -15\)

\(Q = (-15, 4)\)

The tangent meets the radius \(CA\) at right angles, so first find the gradient of \(CA\) from the difference in the \(y\)-coordinates over the difference in the \(x\)-coordinates.

\(\frac{ 3 - (-4) }{ 4 - 2 } = \frac{7}{2}\)

Perpendicular gradients multiply to \(-1\), so the tangent has gradient

\(-1 \div \left(\frac{7}{2}\right) = -\frac{2}{7}\)

The tangent passes through \(A(4, 3)\):

\(y - 3 = -\frac{2}{7}(x - 4)\)

Multiply both sides by 7 to clear the fraction:

\(7(y - 3) = -2(x - 4)\)

\(7y - 21 = -2x + 8\)

Collect every term on one side so that the \(x\) term is positive:

\(2x + 7y - 29 = 0\)

The tangent meets the radius \(OP\) at right angles, so start with the gradient of \(OP\), where \(O\) is the origin.

\(\frac{ 0.5 }{ -1.2 } = -\frac{5}{12}\)

The tangent has the negative reciprocal gradient

\(-1 \div \left(-\frac{5}{12}\right) = \frac{12}{5}\)

The tangent passes through \((-1.2, 0.5)\):

\(y - 0.5 = \frac{12}{5}\left(x + 1.2\right)\)

Multiply both sides by 5 to clear the fraction:

\(5\left(y - 0.5\right) = 12\left(x + 1.2\right)\)

\(5y - 2.5 = 12x + 14.4\)

Collect the \(x\) and \(y\) terms on the left and the numbers on the right. The number on the right is

\(5 \times (0.5) - (12) \times (-1.2) = 16.9\)

so

\(-12x + 5y = 16.9\)

Multiply both sides by 10 to clear the decimal:

\(-120x + 50y = 169\)

The highest common factor of 120, 50 and 169 is 1, so in its simplest integer form the tangent is

\(-120x + 50y = 169\)

Parallel lines have equal gradients, so the tangent at \(P\) has gradient \(\frac{15}{8}\).

The radius \(OP\) meets the tangent at right angles, so

\(-1 \div \frac{15}{8} = -\frac{8}{15}\)

\(OP\) starts at the origin, so \(P\) lies on the line \(y = -\frac{8}{15}x\), and every point of that line can be written as \((t, -\frac{8}{15}t)\) for some \(t\).

\(P\) is also on the circle, so put those coordinates into its equation:

\(t^2 + \left(-\frac{8}{15}t\right)^2 = 289\)

\(t^2 + \frac{ 64 }{ 225 }t^2 = 289\)

\(\frac{ 225 + 64 }{ 225 }t^2 = 289\)

\(\frac{ 289 }{ 225 }t^2 = 289\)

\(t^2 = 225\), so \(t = 15\) or \(t = -15\)

\(P\) is in the second quadrant, where \(x\) is negative, so \(t = -15\).

\(P\) is \((-15, 8)\)

The tangent is perpendicular to the radius \(OP\), so start with the gradient of \(OP\). \(O\) is \((0, 0)\) and \(P\) is \((-8, -4)\).

\(\text{gradient of } OP = \frac{ -4 - 0 }{ -8 - 0 } = \frac{1}{2}\)

Perpendicular lines have gradients that multiply to \(-1\), so the tangent's gradient is the negative reciprocal.

\(\text{gradient of tangent} = -1 \div \frac{1}{2} = -2\)

This is the gradient given in the question.

Use the gradient \(-2\) and the point \(P(-8, -4)\) to write down the equation of the tangent.

\(y - (-4) = -2(x - (-8))\)

\(y = -2x - 20\)

So the tangent crosses the \(y\)-axis at \(B(0, -20)\).

At \(A\), \(y = 0\):

\(0 = -2x - 20\)

\(x = -10\), so \(A\) is \((-10, 0)\).

\(OA = 10\) and \(OB = 20\), and angle \(AOB\) is a right angle, so by Pythagoras:

\(AB^2 = 10^2 + 20^2 = 100 + 400 = 500\)

\(AB = \sqrt{ 500 } = 10\sqrt{ 5 }\)