TMUA: Circles
Circles in coordinate geometry and the circle theorems.
- Section 1 Part 1 · MM3 Coordinate geometry in the (x, y)-plane · spec MM3.2-MM3.3
- Papers 1 and 2
- 8 practice questions
- No calculator
What the specification covers
- Completing the square to find centre and radius
- Tangent perpendicular to radius; chord bisected by the perpendicular from the centre
- Circle theorems used in coordinate problems
Key ideas
- \((x-a)^2+(y-b)^2=r^2\) has centre \((a,b)\) and radius \(r\). Complete the square on \(x^2+y^2+cx+dy+e=0\) to find them.
- The perpendicular from the centre to a chord bisects it: half-chord \(=\sqrt{r^2-d^2}\) where \(d\) is the distance from the centre.
- A tangent is perpendicular to the radius at the point of contact; the tangent length from an outside point is \(\sqrt{(\text{distance to centre})^2-r^2}\).
- Angle at the centre is twice the angle at the circumference; angle in a semicircle is \(90^\circ\); opposite angles of a cyclic quadrilateral add to \(180^\circ\).
Common mistakes
- From \(x^2+y^2-6x+4y-12=0\) the radius is \(\sqrt{9+4+12}=5\): add the squares' constants to the right-hand side.
- Where a circle meets an axis: set the other coordinate to 0 and solve the quadratic.
Exam tip
Draw the right angle (radius and tangent, or centre and chord midpoint) and use Pythagoras.
Worked example
Worked example
The circle \(x^2+y^2-6x+4y-12=0\) cuts the \(y\)-axis at two points. What is the distance between them?
- \(8\)
- \(4\sqrt{3}\)
- \(6\)
- \(10\)
- \(2\sqrt{21}\)
Answer: A: \(8\)
Put \(x=0\): \(y^2+4y-12=0\), \((y+6)(y-2)=0\), so \(y=2\) or \(y=-6\). The distance is \(8\). (Check: centre \((3,-2)\), radius 5; half-chord \(\sqrt{25-9}=4\).)
Practice questions
Try each question before opening the solution. No calculator: the TMUA does not allow one. Every answer here was re-checked independently by computer before it was published.
Question 1
A tangent is drawn from the point \((7,1)\) to the circle \((x-1)^2+(y-1)^2=20\). What is the length of the tangent from \((7,1)\) to its point of contact?
- \(2\sqrt{5}\)
- \(4\)
- \(6\)
- \(16\)
- \(2\sqrt{14}\)
Show the answer and solution
Answer: B: \(4\)
The tangent is perpendicular to the radius at the point of contact, so by Pythagoras (tangent length)\(^2\) = (distance to centre)\(^2\) \(-\,r^2=6^2-20=16\). The length is 4.
Question 2
The circle \(x^2+y^2+6x-8y=0\) cuts the \(y\)-axis at two points. What is the distance between them?
- \(8\)
- \(5\)
- \(\sqrt{91}\)
- \(4\)
- \(10\)
Show the answer and solution
Answer: A: \(8\)
Completing the square, the centre is \((-3,4)\) and the radius is \(5\). The distance from the centre to the \(y\)-axis is \(3\), so the half-chord is \(\sqrt{25-9}=4\) and the chord has length \(8\).
Question 3
The circle \(x^2+y^2-6x+2y-15=0\) cuts the \(y\)-axis at two points. What is the distance between them?
- \(4\)
- \(5\)
- \(8\)
- \(10\)
- \(\sqrt{91}\)
Show the answer and solution
Answer: C: \(8\)
Completing the square, the centre is \((3,-1)\) and the radius is \(5\). The distance from the centre to the \(y\)-axis is \(3\), so the half-chord is \(\sqrt{25-9}=4\) and the chord has length \(8\).
Question 4
The circle \(x^2+y^2+8x+4y-5=0\) cuts the \(y\)-axis at two points. What is the distance between them?
- \(3\)
- \(5\)
- \(10\)
- \(6\)
- \(2 \sqrt{21}\)
Show the answer and solution
Answer: D: \(6\)
Completing the square, the centre is \((-4,-2)\) and the radius is \(5\). The distance from the centre to the \(y\)-axis is \(4\), so the half-chord is \(\sqrt{25-16}=3\) and the chord has length \(6\).
Question 5
The circle \(x^2+y^2+8x+2y-8=0\) cuts the \(y\)-axis at two points. What is the distance between them?
- \(3\)
- \(6\)
- \(10\)
- \(2 \sqrt{21}\)
- \(5\)
Show the answer and solution
Answer: B: \(6\)
Completing the square, the centre is \((-4,-1)\) and the radius is \(5\). The distance from the centre to the \(y\)-axis is \(4\), so the half-chord is \(\sqrt{25-16}=3\) and the chord has length \(6\).
Question 6
The circle \(x^2+y^2+8x-6y=0\) cuts the \(y\)-axis at two points. What is the distance between them?
- \(6\)
- \(2 \sqrt{21}\)
- \(10\)
- \(5\)
- \(3\)
Show the answer and solution
Answer: A: \(6\)
Completing the square, the centre is \((-4,3)\) and the radius is \(5\). The distance from the centre to the \(y\)-axis is \(4\), so the half-chord is \(\sqrt{25-16}=3\) and the chord has length \(6\).
Question 7
The circle \(x^2+y^2+8x-8y+7=0\) cuts the \(y\)-axis at two points. What is the distance between them?
- \(2 \sqrt{21}\)
- \(10\)
- \(3\)
- \(5\)
- \(6\)
Show the answer and solution
Answer: E: \(6\)
Completing the square, the centre is \((-4,4)\) and the radius is \(5\). The distance from the centre to the \(y\)-axis is \(4\), so the half-chord is \(\sqrt{25-16}=3\) and the chord has length \(6\).
Keep going
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