Coordinate Geometry of the Circle

The circle $C$ has equation $x^2 + y^2 - 4x + 2y + k = 0$, where $k$ is a constant. $C$ has radius $3$. What is the value of $k$?

• $k = -4$

  Correct. Completing the square: $(x-2)^2 + (y+1)^2 - 4 - 1 + k = 0$, so $(x-2)^2+(y+1)^2 = 5-k$. Setting the radius to 3 means $5-k=9$, so $k=-4$.

• B$k = 9$

  This treats $k$ as if it were $r^2$ directly, ignoring the $g^2+f^2$ correction the completed square adds. The right-hand side after completing the square is $5-k$, not $-k$ — the $5$ comes from the two halved-and-squared coefficients and cannot be dropped.

• C$k = 4$

  This solves $k - 5 = 9$ instead of $5 - k = 9$ — the sign of the completed-square correction has been reversed. Completing the square turns $+k$ into a term that gets subtracted alongside $4$ and $1$, not added to them.

• D$k = -9$

  This solves $-k = 9$, skipping the $g^2+f^2$ correction entirely (the same missing step as option B, with the opposite sign error layered on top). The full equation to solve is $5-k=9$, not $-k=9$.

Traps tested: G squared plus f squared correction dropped · Sign of constant term reversed

The circle $C$ has equation $x^2 + y^2 + 10x - 4y - 7 = 0$. What are the coordinates of its centre?

• $(-5, 2)$

  Correct. $x^2+10x = (x+5)^2-25$ and $y^2-4y=(y-2)^2-4$, giving $(x+5)^2+(y-2)^2-25-4-7=0$, i.e. $(x+5)^2+(y-2)^2=36$. The centre is $(-5,2)$: the sign in each bracket is flipped to read off the coordinate.

• B$(5, -2)$

  Both signs are read straight off the brackets without being flipped. $(x+5)^2$ is zero at $x=-5$ and $(y-2)^2$ is zero at $y=+2$ — this is the reverse of both.

• C$(5, 2)$

  The x-coordinate's sign has not been flipped; the y-coordinate's has. $(x+5)^2$ being zero at $x=-5$ applies the same way for both variables — there's no reason for one sign to flip and not the other.

• D$(-10, 4)$

  This uses the equation's own coefficients ($10$ and $-4$) directly as centre coordinates, without halving them first. The coefficient of $x$ is always $2 \times$ (the negative of the centre's x-coordinate) — it has to be halved before it becomes a coordinate at all.

Traps tested: Both centre signs flipped · Centre signs not flipped · Coefficient not halved before reading centre

The circle $C$ has equation $(x-2)^2 + (y+3)^2 = 20$. Give the radius of $C$ exactly, in simplified surd form.

• $2\sqrt{5}$

  Correct. $\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}$. The general marking guidance's exact-answer rule expects a value like this to be left in exact form, not converted to a decimal.

• B$4.47$

  This is a correct decimal approximation and still loses the mark: where working with surds is clearly involved, the general marking guidance states marks are normally lost for resorting to a rounded decimal instead of the exact value.

• C$20$

  This is $r^2$, the number on the right-hand side of the equation — the square root has not been taken at all. The radius is the square root of that number, not the number itself.

• D$10$

  This halves $20$ instead of taking its square root. Halving and square-rooting are different operations, and only one of them undoes the squaring that produced $r^2 = 20$ in the first place.

Traps tested: Decimal given when exact form required · R squared mistaken for r · Halved instead of square rooted

A chord of a circle has midpoint $M(3, 4)$. The circle's centre is at $(7, 1)$. What is the gradient of the chord?

• $\frac{4}{3}$

  Correct. The gradient from the centre to $M$ is $\frac{4-1}{3-7} = -\frac{3}{4}$. The perpendicular from the centre bisects the chord, so the chord's own gradient is the negative reciprocal of that: $\frac{4}{3}$.

• B$-\frac{3}{4}$

  This is the gradient of the line from the centre to $M$ itself, used directly as the chord's gradient. That line and the chord are perpendicular to each other, not parallel — the property is specifically that they meet at a right angle.

• C$-\frac{4}{3}$

  This has flipped the sign of the centre-to-$M$ gradient without taking its reciprocal. Perpendicular gradients multiply to $-1$: $-\frac{3}{4} \times \left(-\frac{4}{3}\right) = 1$, not $-1$.

• D$\frac{3}{4}$

  This has taken the reciprocal without flipping the sign. $-\frac{3}{4} \times \frac{3}{4} = -\frac{9}{16}$, not $-1$ — both the reciprocal and the sign flip are needed together.

Traps tested: Chord gradient not perpendicular to bisector · Perpendicular gradient not reciprocated · Perpendicular gradient sign not flipped

$A$ and $B$ are the endpoints of a diameter of a circle. $P$ is another point on the circle. The gradient of $AP$ is $\frac{2}{5}$. What must the gradient of $BP$ be?

• $-\frac{5}{2}$

  Correct. Since $AB$ is a diameter, the angle at $P$ between $AP$ and $BP$ is always $90°$, whichever point on the circle $P$ actually is — so their gradients multiply to $-1$, making $BP$'s gradient the negative reciprocal of $AP$'s: $-\frac{5}{2}$.

• B$\frac{5}{2}$

  The reciprocal is right; the sign is not. $\frac{2}{5} \times \frac{5}{2} = 1$, not $-1$ — perpendicular lines need both the reciprocal AND the sign flip together.

• C$\frac{2}{5}$

  This assumes $AP$ and $BP$ have equal gradients, which would mean $A$, $P$ and $B$ are all on the same straight line — but $P$ is a separate point on the circle, and the semicircle property is specifically about the angle between two DIFFERENT lines meeting at a right angle, not about them coinciding.

• D$-\frac{2}{5}$

  This flips the sign but keeps the same fraction instead of its reciprocal. $\frac{2}{5} \times \left(-\frac{2}{5}\right) = -\frac{4}{25}$, not $-1$.

Traps tested: Perpendicular gradient sign not flipped · Assumes equal gradients instead of perpendicular · Perpendicular gradient not reciprocated

To find where the line $x - 2y - 8 = 0$ meets a circle, one route substitutes $y = \frac{x-8}{2}$ into the circle's equation, and another substitutes $x = 2y+8$. Which statement about the two routes is correct?

• Both reach the same answer for the same marks, but substituting $x = 2y+8$ avoids introducing a fraction, since it is the smaller coefficient being isolated

  Correct. Mark schemes credit any valid method reaching the correct answer — there is no house method here — but the two routes are not equally quick. Isolating $x$ keeps every substituted term as a whole number; isolating $y$ divides by 2 immediately and forces an extra step later to clear the resulting fraction.

• BOnly substituting $y = \frac{x-8}{2}$ is a valid method; $x$ must always be the subject when solving simultaneously with a circle

  There is no such rule. Either variable can be made the subject of the line's equation before substituting into the circle — the general principles for marking explicitly credit any correct method, and the direction is a matter of efficiency, not validity.

• CThe two routes give different answers, because substituting in a different order changes which points the line actually meets

  The line and the circle are fixed before any substitution happens — how the resulting equation is solved cannot change which geometric points they actually meet. Both routes must, and do, reach the identical pair of points.

• DNeither route is safe; the only reliable method is reading the intersection points off a graph

  A sketch is genuinely useful for checking an answer looks sensible, but it cannot produce the exact coordinates a mark scheme requires — both algebraic substitution routes are reliable and each reaches the exact answer directly.

Traps tested: Assumes only one substitution direction is valid · Believes method choice changes the answer · Overclaims that algebra is unreliable