P3.2Circles

(a) Answer: \((6, 2)\)

• M1 for an attempt to complete the square, e.g. \((x - 6)^2 + (y - 2)^2 = \ldots\)
• A1 for \((6, 2)\)

Worked solution: \((x - 6)^2 - 36 + (y - 2)^2 - 4 + 15 = 0\)
\((x - 6)^2 + (y - 2)^2 = 25\), so the centre is \((6, 2)\).

(b) Answer: \(5\)

• M1 for \(r^2 = 6^2 + 2^2 - 15\) (\(= 25\)) using their completed square
• A1 for \(r = 5\)

Worked solution: \(r^2 = 25\), so \(r = 5\).

(a) Answer: \((7, -6)\)

• M1 for an attempt to complete the square, e.g. \((x - 7)^2 + (y + 6)^2 = \ldots\)
• A1 for \((7, -6)\)

Worked solution: \((x - 7)^2 - 49 + (y + 6)^2 - 36 + 46 = 0\)
\((x - 7)^2 + (y + 6)^2 = 39\), so the centre is \((7, -6)\).

(b) Answer: \(\sqrt{39}\)

• M1 for \(r^2 = 7^2 + (-6)^2 - 46\) (\(= 39\)) using their completed square
• A1 for \(r = \sqrt{39}\)

Worked solution: \(r^2 = 39\), so \(r = \sqrt{39}\).

(c) Answer: \(PQ = 2\sqrt{3}\)

• M1 for substituting \(y = 0\) into the equation of \(C\) (or using Pythagoras with the perpendicular from the centre to the chord)
• M1 for solving to obtain \(x = 7 \pm \sqrt{3}\) (or half-chord \(= \sqrt{3}\))
• A1 for \(2\sqrt{3}\)

Worked solution: The perpendicular distance from the centre to the \(x\)-axis is \(6\).
Half the chord \(= \sqrt{39 - 36} = \sqrt{3}\), so \(PQ = 2\sqrt{3}\).
(Or: \(y = 0\) gives \((x - 7)^2 = 3\), so \(x = 7 \pm \sqrt{3}\).)

(a) Answer: \(r^2 = k^2 + (2k)^2 = 5k^2\); \((x - k)^2 + (y - 2k)^2 = 5k^2\) expands to the given equation.

• M1 for the radius from the centre to \(O\): \(r^2 = k^2 + 4k^2 = 5k^2\)
• A1* for expanding \((x - k)^2 + (y - 2k)^2 = 5k^2\) to \(x^2 + y^2 - 2kx - 4ky = 0\)

Worked solution: The radius is the distance from \((k, 2k)\) to \(O\): \(r^2 = k^2 + 4k^2 = 5k^2\).
\((x - k)^2 + (y - 2k)^2 = 5k^2 \Rightarrow x^2 - 2kx + k^2 + y^2 - 4ky + 4k^2 = 5k^2 \Rightarrow x^2 + y^2 - 2kx - 4ky = 0\).

(b) Answer: The radius to \(O\) has gradient \(\frac{2k}{k} = 2\) for every \(k\), so the tangent at \(O\) has gradient \(-\frac{1}{2}\): \(x + 2y = 0\).

• M1 for the gradient of the radius \(O\) to \((k, 2k)\): \(\frac{2k}{k} = 2\), independent of \(k\)
• M1 for the tangent being perpendicular to the radius: gradient \(-\frac{1}{2}\)
• A1 for \(y = -\frac{1}{2}x\) (or \(x + 2y = 0\)) with a statement that it does not depend on \(k\)

Worked solution: The radius from \(O\) to the centre \((k, 2k)\) has gradient \(\dfrac{2k}{k} = 2\), whatever the value of \(k\).
The tangent at \(O\) is perpendicular to this radius, so its gradient is \(-\frac{1}{2}\), and it passes through \(O\): \(y = -\frac{1}{2}x\), i.e. \(x + 2y = 0\), the same for every \(k\).

(c) Answer: \(k = \dfrac{-5 \pm 5\sqrt{5}}{2}\)

• M1 for the distance from the centre to the line: \(|10 - k|\)
• M1 for equating to the radius: \((10 - k)^2 = 5k^2\)
• M1 for solving the quadratic \(k^2 + 5k - 25 = 0\)
• A1 for \(k = \frac{-5 \pm 5\sqrt{5}}{2}\)

Worked solution: The line \(x = 10\) is a tangent when its distance from the centre equals the radius: \(|10 - k| = \sqrt{5}|k|\).
Squaring: \(100 - 20k + k^2 = 5k^2\), so \(4k^2 + 20k - 100 = 0\), i.e. \(k^2 + 5k - 25 = 0\).
\(k = \dfrac{-5 \pm \sqrt{25 + 100}}{2} = \dfrac{-5 \pm 5\sqrt{5}}{2}\).