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Circles and circle theorems — solutions

17 full solutions · spec MM3.2, MM3.3, M5.8, M5.9

Section 1 Part 1: AS-level pure mathematics17 questions
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Every question on this topic with its answer, the working, and a full method: what the question is testing, the reason for each step, the mistakes that are easy to make and what they would give you, and the principle to take away.

1. CIR-01Circles and circle theorems · spec MM3.2 · Paper 1 style · ★☆☆

Find the radius of the circle \(x^2 + y^2 - 6x + 4y - 12 = 0\).

  1. A\(4\)
  2. B\(5\)
  3. C\(1\)
  4. D\(\sqrt{12}\)
  5. E\(\sqrt{13}\)
  6. F\(25\)
Answer: B. \((x-3)^2 + (y+2)^2 = 12 + 9 + 4 = 25\), so \(r = 5\).
Full method — how and why
The expanded form of a circle. Complete the square in \(x\) and in \(y\) to recover the centre-radius form.
  1. Group the terms: \((x^2 - 6x) + (y^2 + 4y) = 12\).
    Move the constant to the right first so the completed squares have somewhere to go.
  2. Complete each square: \(x^2 - 6x = (x-3)^2 - 9\) and \(y^2 + 4y = (y+2)^2 - 4\).
    Halve each coefficient (\(-6 \to -3\), \(4 \to 2\)) and subtract its square.
  3. Substitute: \((x-3)^2 + (y+2)^2 - 13 = 12\), so \((x-3)^2 + (y+2)^2 = 25\).
    The \(-9\) and \(-4\) move across to the right, adding to the constant.
  4. Read off \(r^2 = 25\), so \(r = 5\).
    The right-hand side is \(r^2\), not \(r\).
Where it goes wrong
  • Answering \(25\) - that is \(r^2\).
  • Subtracting the \(9\) and \(4\) on the right instead of adding, giving \(r^2 = 12\) or similar.
Takeaway \(x^2+y^2+cx+dy+e=0\) has centre \(\left(-\frac c2, -\frac d2\right)\) and \(r^2 = \frac{c^2}{4}+\frac{d^2}{4}-e\).
2. CIR-03Circles and circle theorems · spec MM3.2 · Paper 1 style · ★☆☆

The circle with centre \((2, -1)\) passes through \((5, 3)\). Find the equation of the tangent to the circle at \((5, 3)\).

  1. A\(4x - 3y = 11\)
  2. B\(3x + 4y = 27\)
  3. C\(x + y = 8\)
  4. D\(3x + 4y = 15\)
  5. E\(3x - 4y = 3\)
  6. F\(4x + 3y = 29\)
Answer: B. Radius direction \((3, 4)\) has gradient \(\frac43\), so the tangent has gradient \(-\frac34\): \(y - 3 = -\frac34(x - 5) \Rightarrow 4y - 12 = -3x + 15 \Rightarrow 3x + 4y = 27\).
Full method — how and why
A tangent is perpendicular to the radius at the point of contact. Find the radius gradient, invert, and write the line.
  1. Vector from centre \((2,-1)\) to the point \((5,3)\): \((3, 4)\), so the radius has gradient \(\frac43\).
    Differences of coordinates give the direction; no need for the circle's equation.
  2. The tangent is perpendicular, so its gradient is \(-\dfrac34\).
    Tangent \(\perp\) radius is the key circle fact here.
  3. Line through \((5,3)\): \(y - 3 = -\frac34(x - 5)\).
    The tangent touches at the given point, so use it in the point-gradient form.
  4. Clear fractions: \(4y - 12 = -3x + 15 \Rightarrow 3x + 4y = 27\).
    Multiply by \(4\), then collect.
Where it goes wrong
  • Using the radius gradient \(\frac43\) for the tangent, giving \(4x - 3y = 11\) - an offered option.
  • Arithmetic slip on the constant, giving \(3x + 4y = 15\).
Takeaway Tangent gradient \(= -1 \div\) (radius gradient); the point of contact is on both the circle and the tangent.
3. CIR-04Circles and circle theorems · spec MM3.2 · Paper 1 style · ★☆☆

The points \((1, 7)\) and \((5, -1)\) are the ends of a diameter of a circle. Find the equation of the circle.

  1. A\((x-3)^2 + (y-3)^2 = 40\)
  2. B\((x-2)^2 + (y-4)^2 = 20\)
  3. C\((x-3)^2 + (y-3)^2 = \sqrt{20}\)
  4. D\((x-3)^2 + (y-3)^2 = 20\)
  5. E\((x-3)^2 + (y-3)^2 = 80\)
  6. F\((x+3)^2 + (y+3)^2 = 20\)
Answer: D. Centre is the midpoint \((3, 3)\); radius\(^2\) \(= (3-1)^2 + (3-7)^2 = 4 + 16 = 20\).
Full method — how and why
A diameter's endpoints give the centre (midpoint) and the radius (half the distance) in one move.
  1. Centre \(=\) midpoint \(= \left(\frac{1+5}{2}, \frac{7 + (-1)}{2}\right) = (3, 3)\).
    The centre is the midpoint of any diameter.
  2. Radius\(^2\): distance from the centre to \((1,7)\) squared \(= (3-1)^2 + (3-7)^2 = 4 + 16 = 20\).
    Going from centre to an endpoint is one radius; squaring avoids surds entirely.
  3. Write the equation: \((x-3)^2 + (y-3)^2 = 20\).
    The right-hand side is \(r^2\).
Where it goes wrong
  • Using the full diameter length squared (\(80\)) as \(r^2\).
  • Writing \(r^2 = \sqrt{20}\) - the square root is already undone by squaring.
Takeaway From a diameter: centre is the midpoint, and \(r^2\) is the squared distance from centre to either end.
4. CIR-06Circles and circle theorems · spec MM3.3 · Paper 1 style · ★☆☆

\(A\), \(B\) and \(C\) lie on a circle with centre \(O\). Angle \(AOB = 100^\circ\) and \(C\) lies on the major arc \(AB\). Find angle \(ACB\).

  1. A\(200^\circ\)
  2. B\(100^\circ\)
  3. C\(40^\circ\)
  4. D\(50^\circ\)
  5. E\(130^\circ\)
  6. F\(80^\circ\)
Answer: D. The angle at the centre is twice the angle at the circumference subtended by the same arc: \(ACB = 50^\circ\).
Full method — how and why
The angle at the centre is twice the angle at the circumference on the same arc - provided you use the right arc.
  1. Identify the arc: angle \(AOB = 100^\circ\) at the centre stands on the minor arc \(AB\).
    The reflex angle at the centre (\(260^\circ\)) would correspond to the other arc - which arc you use decides the answer.
  2. \(C\) is on the major arc, so it sees the same minor arc \(AB\).
    The circumference angle must be on the opposite arc from the one it stands on.
  3. Apply the theorem: angle \(ACB = \frac12 \times 100^\circ = 50^\circ\).
    Centre angle is double the circumference angle on the same arc.
Where it goes wrong
  • Doubling instead of halving, giving \(200^\circ\) - impossible for an angle in a triangle.
  • Using the reflex angle \(260^\circ\) and answering \(130^\circ\) (which would be right if \(C\) were on the minor arc).
Takeaway Angle at centre \(= 2 \times\) angle at circumference on the same arc; check which side of the chord the point lies.
5. CIR-07Circles and circle theorems · spec MM3.3 · Paper 1 style · ★☆☆

\(PQRS\) is a cyclic quadrilateral with angle \(P = 3x\), angle \(R = x + 40^\circ\). Find \(x\).

  1. A\(45^\circ\)
  2. B\(50^\circ\)
  3. C\(35^\circ\)
  4. D\(30^\circ\)
  5. E\(40^\circ\)
  6. F\(25^\circ\)
Answer: C. Opposite angles of a cyclic quadrilateral sum to \(180^\circ\): \(4x + 40 = 180\), \(x = 35^\circ\).
Full method — how and why
Cyclic quadrilateral: opposite angles are supplementary. That gives a linear equation immediately.
  1. Identify the opposite pair: in \(PQRS\) the vertices go round in order, so \(P\) and \(R\) are opposite.
    Opposite means separated by one vertex on each side - reading the labelling order matters.
  2. Apply the theorem: \(3x + (x + 40) = 180\).
    Opposite angles of a cyclic quadrilateral sum to \(180^\circ\).
  3. Solve: \(4x = 140\), so \(x = 35^\circ\).
    Check: angles \(105^\circ\) and \(75^\circ\) sum to \(180^\circ\) ✓.
Where it goes wrong
  • Setting the angles equal (that is a parallelogram property, not a cyclic one).
  • Using \(360^\circ\) and getting \(x = 80^\circ\).
Takeaway Cyclic quadrilateral: opposite angles sum to \(180^\circ\); adjacent ones have no such relation.
6. CIR-08Circles and circle theorems · spec MM3.3 · Paper 1 style · ★☆☆

A chord of a circle of radius \(13\) is at perpendicular distance \(5\) from the centre. Find the length of the chord.

  1. A\(26\)
  2. B\(\sqrt{144}\)
  3. C\(18\)
  4. D\(12\)
  5. E\(10\)
  6. F\(24\)
Answer: F. The perpendicular from the centre bisects the chord: half-length \(= \sqrt{13^2 - 5^2} = 12\), so the chord is \(24\).
Full method — how and why
The perpendicular from the centre bisects the chord, creating a right-angled triangle with the radius as hypotenuse.
  1. Draw the radius to one end of the chord and the perpendicular from the centre: they form a right-angled triangle.
    The perpendicular distance, half the chord and the radius are the three sides.
  2. Apply Pythagoras: half-chord \(= \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\).
    The radius \(13\) is the hypotenuse because it is the longest side.
  3. Double it: the chord is \(24\).
    The perpendicular bisects the chord, so the triangle gives only half of it.
Where it goes wrong
  • Answering \(12\) by forgetting to double.
  • Adding instead of subtracting inside the square root.
Takeaway Perpendicular from centre to chord: it bisects the chord, so use Pythagoras and then double.
7. CIR-10Circles and circle theorems · spec MM3.3 · Paper 1 style · ★☆☆

\(TA\) is a tangent to a circle at \(A\), and \(AB\) is a chord. \(C\) is a point on the circle on the opposite side of \(AB\) from \(T\), with angle \(ACB = 65^\circ\). Find angle \(TAB\).

  1. A\(25^\circ\)
  2. B\(65^\circ\)
  3. C\(115^\circ\)
  4. D\(130^\circ\)
  5. E\(50^\circ\)
  6. F\(32.5^\circ\)
Answer: B. Alternate segment theorem: the angle between tangent and chord equals the angle in the alternate segment, so \(TAB = 65^\circ\).
Full method — how and why
The alternate segment theorem, stated almost word for word. The work is in identifying which angle is 'in the alternate segment'.
  1. \(TA\) is tangent at \(A\) and \(AB\) is a chord, so the angle \(TAB\) between them is the tangent-chord angle.
    The theorem applies exactly to this configuration.
  2. The alternate segment is the one on the other side of \(AB\) from \(T\) - which is where \(C\) lies.
    The question places \(C\) deliberately; the angle used must be in the segment away from the tangent's side.
  3. Therefore angle \(TAB\) = angle \(ACB = 65^\circ\).
    Alternate segment theorem: the tangent-chord angle equals the inscribed angle in the alternate segment.
Where it goes wrong
  • Using \(180^\circ - 65^\circ = 115^\circ\), which is the angle on the other side of the tangent.
  • Halving to \(32.5^\circ\) by confusing this with the centre-circumference theorem.
Takeaway Tangent-chord angle \(=\) angle in the alternate segment - the segment on the far side of the chord.
8. CIR-12Circles and circle theorems · spec MM3.2 · Paper 1 style · ★☆☆

The circle \(C\) has equation \(x^2 + y^2 - 4x - 6y + 9 = 0\). Which of the following lines does not meet \(C\)?

  1. A\(x = 4\)
  2. B\(x = 2\)
  3. C\(x = 0\)
  4. D\(y = 0\)
  5. E\(y = 5\)
Answer: D. Centre \((2, 3)\), radius \(2\), so the circle spans \(0 \leq x \leq 4\) and \(1 \leq y \leq 5\). The line \(y = 0\) is \(3\) from the centre, more than the radius, so it misses; \(x = 0\), \(x = 4\) and \(y = 5\) touch and \(x = 2\) passes through the centre.
Full method — how and why
Which line misses the circle: compare each line's distance from the centre with the radius. Horizontal and vertical lines make this trivial.
  1. Complete the square: \((x-2)^2 + (y-3)^2 = -9 + 4 + 9 = 4\), so centre \((2,3)\) and radius \(2\).
    Always convert to centre-radius form first.
  2. The circle therefore spans \(x\) from \(0\) to \(4\) and \(y\) from \(1\) to \(5\).
    Centre \(\pm\) radius in each direction gives the bounding box, whose edges are tangent lines.
  3. Check each line: \(x = 0\), \(x = 4\) and \(y = 5\) are tangents (distance exactly \(2\)); \(x = 2\) passes through the centre.
    For a vertical line \(x = k\), the distance from the centre is \(|k - 2|\).
  4. \(y = 0\) is at distance \(|0 - 3| = 3 > 2\), so it misses the circle.
    Distance greater than the radius means no intersection.
Where it goes wrong
  • Assuming a tangent line 'does not meet' the circle - touching counts as meeting.
  • Getting \(r^2 = 9\) by mishandling the \(+9\) constant.
Takeaway For horizontal/vertical lines just compare the centre coordinate with the radius; tangency still counts as meeting.
9. CIR-13Circles and circle theorems · spec MM3.2 · Paper 1 style · ★☆☆

The circle \(x^2 + y^2 = r^2\) and the line \(3x + 4y = 20\) have exactly one point in common. Find \(r\).

  1. A\(\dfrac{20}{7}\)
  2. B\(4\)
  3. C\(20\)
  4. D\(5\)
  5. E\(\sqrt{20}\)
  6. F\(2\)
Answer: B. Distance from origin to the line \(= \frac{20}{\sqrt{9 + 16}} = 4\). Tangency means \(r = 4\).
Full method — how and why
One common point means tangency, so the radius equals the distance from the centre to the line.
  1. The circle is centred at the origin, so compute the distance from \((0,0)\) to \(3x + 4y = 20\).
    Centre-at-origin makes the distance formula especially clean.
  2. Distance \(= \dfrac{|3(0) + 4(0) - 20|}{\sqrt{3^2+4^2}} = \dfrac{20}{5} = 4\).
    \(\sqrt{9+16} = 5\) - the familiar 3-4-5 triple.
  3. Tangency requires \(r\) equal to this distance: \(r = 4\).
    If \(r\) were larger the line would cut the circle twice; smaller, and it would miss.
Where it goes wrong
  • Dividing by \(\sqrt{20}\) or by \(7\) instead of \(5\).
  • Answering \(20\) or \(\sqrt{20}\) by stopping too early.
Takeaway Distance from the origin to \(ax+by=c\) is \(\frac{|c|}{\sqrt{a^2+b^2}}\); tangency sets it equal to \(r\).
10. CIR-14Circles and circle theorems · spec MM3.3 · Paper 1 style · ★☆☆

\(AB\) is a diameter of a circle and \(C\) is a point on the circle with \(AC = 6\) and \(BC = 8\). Find the radius of the circle.

  1. A\(\sqrt{50}\)
  2. B\(4\)
  3. C\(10\)
  4. D\(7\)
  5. E\(5\)
  6. F\(14\)
Answer: E. The angle in a semicircle is a right angle, so \(AB = \sqrt{36 + 64} = 10\) and the radius is \(5\).
Full method — how and why
The angle in a semicircle is a right angle, so \(AB\) is the hypotenuse of a right-angled triangle.
  1. \(AB\) is a diameter and \(C\) is on the circle, so angle \(ACB = 90^\circ\).
    The angle in a semicircle is a right angle - the key theorem here.
  2. Apply Pythagoras: \(AB = \sqrt{6^2 + 8^2} = \sqrt{100} = 10\).
    The right angle is at \(C\), so the diameter \(AB\) is the hypotenuse.
  3. Radius \(= \frac{10}{2} = 5\).
    The question asks for the radius, not the diameter.
Where it goes wrong
  • Answering \(10\) (the diameter).
  • Assuming the right angle is at \(A\) or \(B\).
Takeaway A triangle inscribed with one side a diameter is right-angled at the third vertex.
11. CIR-15Circles and circle theorems · spec MM3.2 · Paper 1 style · ★☆☆

A circle passes through \((0, 0)\), \((6, 0)\) and \((0, 8)\). Find the coordinates of its centre.

  1. A\((5, 5)\)
  2. B\((2, 3)\)
  3. C\((4, 3)\)
  4. D\((6, 8)\)
  5. E\((3, 4)\)
  6. F\((3, 3)\)
Answer: E. The triangle has a right angle at the origin, so \((6, 0)\)-\((0, 8)\) is a diameter; the centre is its midpoint \((3, 4)\).
Full method — how and why
Three points determine a circle, but here two of the chords are along the axes - and the right angle at the origin gives the centre instantly.
  1. Notice the triangle \((0,0)\), \((6,0)\), \((0,8)\) has a right angle at the origin.
    The two legs lie along the axes, so the angle between them is \(90^\circ\).
  2. A right angle inscribed in a circle stands on a diameter, so the hypotenuse from \((6,0)\) to \((0,8)\) is a diameter.
    This is the converse of the angle-in-a-semicircle theorem.
  3. The centre is the midpoint of that diameter: \(\left(\frac{6+0}{2}, \frac{0+8}{2}\right) = (3, 4)\).
    Centre is the midpoint of any diameter.
Where it goes wrong
  • Averaging all three points to get \((2, \frac83)\) - that is the centroid, not the circumcentre.
  • Taking the midpoint of the wrong pair, e.g. \((3,0)\).
Takeaway If three given points form a right angle, the hypotenuse is a diameter and its midpoint is the centre.
12. X2-10Circles and circle theorems · spec MM3.2 · Paper 2 style · ★☆☆

The circle \(C\) has centre \((a, b)\) and radius \(r\). \(P\): '\(r > |b|\)'. \(Q\): '\(C\) crosses the \(x\)-axis at two distinct points'. Which one of the following is true?

  1. AP is necessary and sufficient for Q.
  2. BP is necessary but not sufficient for Q.
  3. CP is sufficient but not necessary for Q.
  4. DP is neither necessary nor sufficient for Q.
Answer: A. The distance from the centre to the \(x\)-axis is \(|b|\); the circle cuts the axis twice exactly when this is less than \(r\).
Full method — how and why
Translate the geometry into a distance comparison: circle versus line depends on centre-to-line distance against the radius.
  1. The perpendicular distance from the centre \((a,b)\) to the \(x\)-axis is \(|b|\).
    The \(x\)-axis is \(y = 0\), so the vertical distance is the size of the \(y\)-coordinate.
  2. If \(|b| < r\) the circle reaches across the axis and cuts it twice; if \(|b| = r\) it touches once; if \(|b| > r\) it misses entirely.
    Three cases, decided by comparing distance with radius.
  3. So 'two distinct points' happens exactly when \(r > |b|\) - the two statements are equivalent.
    Necessary and sufficient.
Where it goes wrong
  • Writing the condition as \(r > b\), which fails for a centre below the axis (\(b\) negative). The modulus matters.
  • Including the tangent case \(r = |b|\), which gives one point, not two.
Takeaway Circle versus line: compare the perpendicular distance from the centre with \(r\) - less gives two points, equal one, greater none.
13. CIR-02Circles and circle theorems · spec MM3.2 · Paper 1 style · ★★☆

Find the complete set of values of \(k\) for which \(x^2 + y^2 + 2kx - 4y + 3k + 4 = 0\) represents a circle.

  1. A\(k > 3\)
  2. B\(0 < k < 3\)
  3. C\(k < -3\) or \(k > 0\)
  4. D\(k < 0\) or \(k > 3\)
  5. E\(-3 < k < 0\)
  6. Fall real \(k\)
Answer: D. \((x + k)^2 + (y - 2)^2 = k^2 + 4 - 3k - 4 = k^2 - 3k = k(k - 3)\). A circle needs a positive right-hand side: \(k < 0\) or \(k > 3\).
Full method — how and why
An equation of circle form only is a circle when the resulting \(r^2\) is positive. That gives an inequality in \(k\).
  1. Complete both squares: \((x + k)^2 - k^2 + (y - 2)^2 - 4 + 3k + 4 = 0\).
    Treat \(k\) as a constant: half of \(2k\) is \(k\), half of \(-4\) is \(-2\).
  2. Rearrange: \((x+k)^2 + (y-2)^2 = k^2 - 3k\).
    The \(-4\) and \(+4\) cancel, leaving \(r^2 = k^2 - 3k\).
  3. A genuine circle needs \(r^2 > 0\): \(k^2 - 3k > 0\), i.e. \(k(k - 3) > 0\).
    If \(r^2 = 0\) the 'circle' is a single point; if \(r^2 < 0\) there are no real points at all.
  4. Solve the quadratic inequality: \(k < 0\) or \(k > 3\).
    Roots \(0\) and \(3\), upward parabola, positive outside the roots.
Where it goes wrong
  • Choosing \(0 < k < 3\) - that is where \(r^2\) is negative and there is no circle.
  • Forgetting the sign change when moving \(-k^2\) across.
Takeaway Complete the square and demand \(r^2 > 0\); the shape only exists when the right-hand side is positive.
14. CIR-05Circles and circle theorems · spec MM3.2 · Paper 1 style · ★★☆

The line \(y = x + c\) is a tangent to the circle \((x-1)^2 + (y-2)^2 = 8\). Find the possible values of \(c\).

  1. A\(c = -3\) or \(c = 5\)
  2. B\(c = 1 \pm 2\sqrt{2}\)
  3. C\(c = 1\) or \(c = 5\)
  4. D\(c = -1\) or \(c = 3\)
  5. E\(c = \pm 4\)
  6. F\(c = -5\) or \(c = 3\)
Answer: A. The distance from \((1, 2)\) to \(x - y + c = 0\) is \(\frac{|1 - 2 + c|}{\sqrt2} = \sqrt8 = 2\sqrt2\), so \(|c - 1| = 4\): \(c = 5\) or \(c = -3\).
Full method — how and why
Tangency means the perpendicular distance from the centre to the line equals the radius. Faster than substituting and using the discriminant.
  1. Identify centre \((1,2)\) and radius \(\sqrt8 = 2\sqrt2\).
    Read them straight from the centre-radius form.
  2. Write the line in the form \(x - y + c = 0\) and apply the distance formula: \(\dfrac{|1 - 2 + c|}{\sqrt{1^2 + (-1)^2}} = \dfrac{|c - 1|}{\sqrt2}\).
    Rearranging \(y = x + c\) to \(x - y + c = 0\) gives \(a = 1\), \(b = -1\).
  3. Set it equal to the radius: \(\dfrac{|c-1|}{\sqrt2} = 2\sqrt2\), so \(|c - 1| = 4\).
    Multiply both sides by \(\sqrt2\): \(2\sqrt2 \times \sqrt2 = 4\).
  4. Solve the modulus: \(c - 1 = \pm 4\), so \(c = 5\) or \(c = -3\).
    Two tangents with that gradient exist - one on each side of the circle.
Where it goes wrong
  • Dropping the modulus and finding only \(c = 5\).
  • Using \(r = 8\) instead of \(\sqrt8\).
Takeaway Tangency \(\Leftrightarrow\) distance from centre to line \(=\) radius; the modulus always gives two answers.
15. CIR-09Circles and circle theorems · spec MM3.2 · Paper 1 style · ★★☆

Find the length of the tangent from the point \((7, 1)\) to the circle \(x^2 + y^2 - 2x - 4y - 4 = 0\).

  1. A\(3\)
  2. B\(\sqrt{28}\)
  3. C\(6\)
  4. D\(\sqrt{26}\)
  5. E\(\sqrt{37}\)
  6. F\(5\)
Answer: B. Centre \((1, 2)\), radius \(3\). Distance from \((7, 1)\) to centre: \(\sqrt{36 + 1} = \sqrt{37}\). Tangent length \(= \sqrt{37 - 9} = \sqrt{28}\).
Full method — how and why
The tangent, radius and line to the external point form a right-angled triangle. Pythagoras gives the tangent length.
  1. Put the circle in centre-radius form: \((x-1)^2 + (y-2)^2 = 4 + 1 + 4 = 9\), so centre \((1,2)\), radius \(3\).
    Complete the square in \(x\) and \(y\); the constant \(-4\) moves across.
  2. Distance from \((7,1)\) to the centre: \(\sqrt{(7-1)^2 + (1-2)^2} = \sqrt{36 + 1} = \sqrt{37}\).
    This is the hypotenuse of the right-angled triangle (the tangent meets the radius at \(90^\circ\)).
  3. Tangent length \(= \sqrt{PC^2 - r^2} = \sqrt{37 - 9} = \sqrt{28}\).
    Subtract the squares, not the lengths - the right angle is at the point of contact.
Where it goes wrong
  • Answering \(\sqrt{37}\) (that is the distance to the centre) or \(\sqrt{37} - 3\).
  • Getting \(r = 9\) instead of \(3\) from the completed square.
Takeaway Tangent length from \(P\): \(\sqrt{PC^2 - r^2}\), because tangent \(\perp\) radius makes a right-angled triangle.
16. CIR-11Circles and circle theorems · spec MM3.2 · Paper 2 style · ★★☆

Two circles have equations \(x^2 + y^2 = 25\) and \((x - 8)^2 + y^2 = 9\). Which of the following statements is/are true?

  1. IThe circles touch externally.
  2. IIThe circles intersect at exactly two points.
  3. IIIThe point \((5, 0)\) lies on both circles.
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Answer: F. Radii \(5\) and \(3\), centres \(8\) apart \(= 5 + 3\), so the circles touch externally at one point (I true, II false). That point is \((5, 0)\): \(25 = 25\) and \(9 + 0 = 9\) (III true).
Full method — how and why
Two circles: compare the distance between centres with the sum and difference of the radii.
  1. Read off centres and radii: \((0,0)\) with \(r = 5\); \((8,0)\) with \(r = 3\).
    Both are already in centre-radius form.
  2. Distance between centres: \(8\). Compare with \(r_1 + r_2 = 8\).
    Equal to the sum means the circles touch externally - exactly one common point.
  3. I is therefore true, and II (two intersection points) is false.
    Two intersections would need \(|r_1 - r_2| < d < r_1 + r_2\), i.e. \(2 < d < 8\) strictly.
  4. III: check \((5,0)\) in both: \(25 + 0 = 25\) ✓ and \((5-8)^2 + 0 = 9\) ✓. True.
    The touching point lies on the line joining the centres, \(5\) from one and \(3\) from the other.
  5. I and III only.
    Match to the options.
Where it goes wrong
  • Assuming circles that 'meet' must meet twice.
  • Forgetting to test III in both equations.
Takeaway \(d = r_1 + r_2\): touch externally; \(d = |r_1 - r_2|\): touch internally; strictly between: two points.
17. CIR-16Circles and circle theorems · spec MM3.2 · Paper 1 style · ★★☆

The point \(P(x, y)\) moves so that its distance from \((3, 0)\) is twice its distance from \((0, 0)\). The locus of \(P\) is a circle. Find its radius.

  1. A\(\sqrt{2}\)
  2. B\(3\)
  3. C\(2\)
  4. D\(4\)
  5. E\(\sqrt{3}\)
  6. F\(1\)
Answer: C. \((x-3)^2 + y^2 = 4(x^2 + y^2) \Rightarrow 3x^2 + 3y^2 + 6x - 9 = 0 \Rightarrow x^2 + 2x + y^2 = 3 \Rightarrow (x+1)^2 + y^2 = 4\). Radius \(2\).
Full method — how and why
A locus defined by a distance ratio. Write the condition algebraically, square to remove roots, and complete the square.
  1. Write the condition: \(\sqrt{(x-3)^2 + y^2} = 2\sqrt{x^2 + y^2}\).
    'Distance from \(A\) is twice the distance from \(B\)' - write both distances with the formula.
  2. Square both sides: \((x-3)^2 + y^2 = 4(x^2 + y^2)\).
    Squaring is safe because both sides are non-negative; and it removes both roots at once. Note the right-hand side gains a factor \(4\), not \(2\).
  3. Expand and collect: \(x^2 - 6x + 9 + y^2 = 4x^2 + 4y^2 \Rightarrow 3x^2 + 3y^2 + 6x - 9 = 0\).
    Bring everything to one side.
  4. Divide by \(3\) and complete the square: \(x^2 + 2x + y^2 = 3 \Rightarrow (x+1)^2 + y^2 = 4\), so \(r = 2\).
    Dividing by the common factor first makes the coefficients of \(x^2\) and \(y^2\) equal to \(1\) - necessary before completing the square.
Where it goes wrong
  • Squaring the right side as \(2(x^2+y^2)\) instead of \(4(x^2+y^2)\).
  • Forgetting to divide by \(3\) before completing the square, which corrupts the radius.
Takeaway Distance-ratio loci (Apollonius circles): square the condition, divide so \(x^2\) and \(y^2\) have coefficient \(1\), then complete the square.