P2.2 Coordinate geometry of circles
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Pure Mathematics 2 (AS), specification section 3.1: the equation of a circle, circle properties, tangents and how a line meets a circle.
What this page covers
7 points
- The equation \((x - a)^2 + (y - b)^2 = r^2\) and where it comes from
- The expanded form \(x^2 + y^2 + 2gx + 2fy + c = 0\), and completing the square to find the centre and radius
- Three circle properties: the angle in a semicircle, the perpendicular from the centre to a chord, and the tangent perpendicular to the radius
- Equations of tangents and normals at a point on a circle
- Where a line meets a circle, and using the discriminant to decide between two points, a tangent and no meeting
- The circle through three given points
- The length of a tangent from an outside point, and the equations of the tangents from it
In one minute
6 points
- A circle with centre \((a, b)\) and radius \(r\) has equation \((x - a)^2 + (y - b)^2 = r^2\); watch the signs, because centre \((-2, 1)\) gives \((x + 2)^2 + (y - 1)^2\).
- Given the expanded form, complete the square in \(x\) and in \(y\). The coefficients of \(x^2\) and \(y^2\) must be equal (divide through first if they are not 1).
- Angle in a semicircle = 90°; the perpendicular from the centre bisects a chord; the tangent is perpendicular to the radius at the point of contact.
- Tangent at \(P\): find the gradient of the radius \(CP\), then take the negative reciprocal.
- Line meets circle: substitute to get a quadratic. \(b^2 - 4ac > 0\) two points, \(= 0\) tangent, \(< 0\) no meeting.
- Three points: the perpendicular bisectors of two chords meet at the centre. Tangent length from \(P\): \(PT^2 = PC^2 - r^2\).
Contents
13 sections
- What is the equation of a circle?
- How do you find the centre and radius from the expanded form?
- Which circle properties do you need, and how are they used?
- How do you find the equation of a tangent or normal?
- Where does a line meet a circle, and when is it a tangent?
- How do you find the circle through three points?
- How long is a tangent from a point outside the circle?
- Exam tips for P2.2
- Common misconceptions
- How this topic connects
- Key formulae
- Check your understanding
- Specification checklist
What is the equation of a circle?
A circle is Pythagoras around a fixed point. Put that together with the straight-line toolkit from P1 (gradients, midpoints, perpendicular lines) and the discriminant of a quadratic, and every circle question becomes a short chain of familiar steps.
For a general point \((x, y)\) on the circle, the horizontal and vertical distances from the centre are \(x - a\) and \(y - b\). They are the two shorter sides of a right-angled triangle whose hypotenuse is the radius, so Pythagoras gives the equation directly.
- Centre \((a, b)\), radius \(r > 0\). Centre at the origin: \(x^2 + y^2 = r^2\).
- In the formulae list? Not in the formulae booklet: learn it.
The signs inside the brackets are the opposite of the centre's coordinates: \((x + 2)^2\) means \(a = -2\).
- Inside, on or outside? Substitute a point into \((x - a)^2 + (y - b)^2\): less than \(r^2\) means inside, equal means on the circle, greater means outside.
- Diameter given by its ends \(A\) and \(B\): the centre is the midpoint of \(AB\) and the radius is half of \(AB\).
- Leave \(r^2\) as a number on the right-hand side; there is no need to find \(r\) unless it is asked for.
The circle \(C\) has centre \((-2, 1)\) and passes through the point \(P(4, 9)\).
(a) Find an equation for \(C\). (b) Write it in the form \(x^2 + y^2 + px + qy + s = 0\). (c) Determine whether the point \(Q(5, -5)\) lies inside or outside \(C\). (d) A different circle has diameter \(AB\), where \(A\) is \((1, -3)\) and \(B\) is \((7, 5)\). Find its equation.
(a) The radius is the distance \(CP\): \(r^2 = (4 - (-2))^2 + (9 - 1)^2 = 36 + 64 = 100\), so \((x + 2)^2 + (y - 1)^2 = 100\).
(b) Expand and collect: \(x^2 + 4x + 4 + y^2 - 2y + 1 = 100\), so \(x^2 + y^2 + 4x - 2y - 95 = 0\).
(c) Compare \(CQ^2\) with \(r^2\): \((5 + 2)^2 + (-5 - 1)^2 = 49 + 36 = 85 < 100\), so \(Q\) is inside \(C\).
(d) Centre = midpoint of \(AB\) = \(\left(\frac{1 + 7}{2}, \frac{-3 + 5}{2}\right) = (4, 1)\). \(AB = \sqrt{6^2 + 8^2} = 10\), so \(r = 5\): \((x - 4)^2 + (y - 1)^2 = 25\).
Writing the radius instead of its square on the right: \((x + 2)^2 + (y - 1)^2 = 10\). The right-hand side is \(r^2\); a circle of radius 10 needs 100.
How do you find the centre and radius from the expanded form?
- Expanding \((x - a)^2 + (y - b)^2 = r^2\) always gives \(x^2 + y^2\) + (terms in \(x\) and \(y\)) + constant = 0, with equal coefficients of \(x^2\) and \(y^2\) and no \(xy\) term.
- To go back, complete the square separately in \(x\) and in \(y\): \(x^2 - 6x = (x - 3)^2 - 9\).
- If the coefficients of \(x^2\) and \(y^2\) are both 2 (or 3, …), divide the whole equation by that number first.
- The right-hand side after completing the square must be positive; if it is zero or negative, the equation is not a circle.
- Centre \((-g, -f)\), radius \(\sqrt{g^2 + f^2 - c}\), provided \(g^2 + f^2 - c > 0\).
- In the formulae list? Not in the formulae booklet: derive it by completing the square.
The circle \(C\) has equation \(2x^2 + 2y^2 - 12x + 8y + 1 = 0\).
(a) Find the coordinates of the centre of \(C\) and the exact radius. (b) Find the \(x\)-coordinates of the points where \(C\) meets the \(x\)-axis, giving your answers to 3 significant figures. (c) The equation \(x^2 + y^2 + 6x - 8y + k = 0\) represents a circle. Find the set of possible values of the constant \(k\).
(a) Divide by 2, then complete the square in each variable: \(x^2 + y^2 - 6x + 4y + \frac12 = 0\), so \((x - 3)^2 - 9 + (y + 2)^2 - 4 + \frac12 = 0\).
\((x - 3)^2 + (y + 2)^2 = \frac{25}{2}\): centre \((3, -2)\), radius \(\sqrt{12.5} = \frac{5\sqrt2}{2}\) (≈ 3.54).
(b) On the \(x\)-axis, \(y = 0\): \((x - 3)^2 + 4 = 12.5\), so \((x - 3)^2 = 8.5\) and \(x = 3 \pm \sqrt{8.5}\): \(x = 0.0845\) and \(x = 5.92\) (3 s.f.).
(c) Complete the square: \((x + 3)^2 + (y - 4)^2 = 25 - k\). A circle needs \(25 - k > 0\), so \(k < 25\).
Show the completed-square line, \((x - 3)^2 + (y + 2)^2 = 12.5\), before you state the centre and radius. Quoting \((-g, -f)\) from memory leaves no method visible if a sign slips.
Which circle properties do you need, and how are they used?
Three geometric facts do most of the work. In coordinates each one becomes a statement about gradients or distances, which you already know how to calculate.
| Property | In coordinates | Typical use |
|---|---|---|
| The angle in a semicircle is a right angle | if \(AD\) is a diameter and \(B\) is on the circle, gradient \(AB\) × gradient \(BD = -1\) | showing a chord is a diameter; finding the centre as a midpoint |
| The perpendicular from the centre to a chord bisects the chord | the centre lies on the perpendicular bisector of every chord | finding the centre; distances from the centre to a chord |
| The tangent at a point is perpendicular to the radius there | gradient of tangent × gradient of radius \(= -1\) | tangent equations; right-angled triangles for tangent lengths |
- The converse of the first fact is just as useful: if \(\angle ABD = 90°\), then \(A\), \(B\) and \(D\) lie on a circle with \(AD\) as a diameter.
- The perpendicular from the centre \(C\) to a chord, half the chord and a radius form a right-angled triangle: (half-chord)² + (distance from \(C\))² = \(r^2\).
The points \(A(-3, 1)\), \(B(6, 4)\) and \(D(7, 1)\) are given.
(a) Show that \(\angle ABD = 90°\). (b) Hence find an equation of the circle through \(A\), \(B\) and \(D\). (c) \(M\) is the midpoint of \(BD\) and \(C\) is the centre of the circle. Verify that \(CM\) is perpendicular to \(BD\), and find the exact length \(CM\).
(a) Gradient \(AB = \frac{4 - 1}{6 - (-3)} = \frac13\); gradient \(BD = \frac{1 - 4}{7 - 6} = -3\); product \(= -1\), so \(\angle ABD = 90°\).
(b) The angle in a semicircle is 90°, so \(AD\) is a diameter. Centre = midpoint of \(AD\) = \((2, 1)\); radius = \(\frac12 \times 10 = 5\): \((x - 2)^2 + (y - 1)^2 = 25\).
(c) \(M = (6.5, 2.5)\). Gradient \(CM = \frac{2.5 - 1}{6.5 - 2} = \frac13\), and \(\frac13 \times (-3) = -1\), so \(CM \perp BD\). ✓
\(CM^2 = 4.5^2 + 1.5^2 = 22.5\), so \(CM = \frac{3\sqrt{10}}{2}\) (≈ 4.74).
Check with Pythagoras: \(BM^2 = 0.5^2 + 1.5^2 = 2.5\), and \(22.5 + 2.5 = 25 = r^2\). ✓
Assuming a chord is a diameter because it looks long. Only a right angle subtended at the circle (or the chord passing through the centre) proves it. Show the gradient product \(-1\) first, then write "so \(AD\) is a diameter".
How do you find the equation of a tangent or normal?
- Check the point \(P\) really lies on the circle (substitute it).
- Find the gradient of the radius \(CP\): \(m_r = \dfrac{y_P - b}{x_P - a}\).
- Tangent: gradient \(-\dfrac{1}{m_r}\), through \(P\): \(y - y_P = m(x - x_P)\).
The normal at \(P\) is the line through \(P\) perpendicular to the tangent. For a circle it is the line \(CP\), so it always passes through the centre. If the radius is horizontal the tangent is vertical (\(x\) = constant), and vice versa, so no gradient calculation is needed.
The circle \(C\) has equation \((x - 1)^2 + (y + 2)^2 = 20\). The point \(P(3, 2)\) lies on \(C\).
(a) Find an equation of the tangent to \(C\) at \(P\), in the form \(ax + by + c = 0\) where \(a\), \(b\) and \(c\) are integers. (b) Find an equation of the normal to \(C\) at \(P\). (c) The tangent meets the coordinate axes at \(R\) and \(S\). Find the area of triangle \(ORS\), where \(O\) is the origin.
Check \(P\) is on \(C\): \((3 - 1)^2 + (2 + 2)^2 = 4 + 16 = 20\). ✓
(a) Gradient of the radius from the centre \((1, -2)\): \(m_r = \frac{2 - (-2)}{3 - 1} = 2\), so the tangent gradient is \(-\frac12\).
\(y - 2 = -\frac12(x - 3)\), so \(2y - 4 = -x + 3\), giving \(x + 2y - 7 = 0\).
(b) The normal is the line through \(P\) with gradient 2: \(y - 2 = 2(x - 3)\), so \(y = 2x - 4\) (check: at \(x = 1\), \(y = -2\), the centre ✓).
(c) Intercepts of \(x + 2y = 7\): \(R(7, 0)\), \(S(0, 3.5)\). Area \(= \frac12 \times 7 \times 3.5 =\) \(\frac{49}{4} = 12.25\).
When the answer must be "in the form \(ax + by + c = 0\) with integers", clear the fraction in the gradient before you rearrange (multiply by 2 here). Leaving \(y = -\frac12 x + 3.5\) does not meet the request.
Where does a line meet a circle, and when is it a tangent?
Make \(y\) (or \(x\)) the subject of the line's equation and substitute into the circle. The result is a quadratic in one variable. Solve it for the coordinates of the intersection points, then use the line to find the other coordinate. Without solving, the discriminant of that quadratic tells you how many points there are.
| Discriminant | Roots | Line and circle |
|---|---|---|
| \(b^2 - 4ac > 0\) | two distinct real roots | line cuts the circle at two points (a chord) |
| \(b^2 - 4ac = 0\) | one repeated root | line is a tangent |
| \(b^2 - 4ac < 0\) | no real roots | line misses the circle |
When the line contains an unknown constant (\(y = x + k\), or \(y = mx + 3\)), the discriminant gives an equation or inequality for that constant.
The circle \(C\) has equation \(x^2 + y^2 - 4x - 6y + 5 = 0\). The line \(l\) has equation \(y = x + k\), where \(k\) is a constant.
(a) Show that the \(x\)-coordinates of any points where \(l\) meets \(C\) satisfy \(2x^2 + (2k - 10)x + k^2 - 6k + 5 = 0\). (b) Find the values of \(k\) for which \(l\) is a tangent to \(C\), and the point of contact in each case. (c) Find the set of values of \(k\) for which \(l\) meets \(C\) at two distinct points. (d) For \(k = 1\), find the length of the chord cut off by \(C\).
Centre–radius form: \((x - 2)^2 + (y - 3)^2 = 8\), so centre \((2, 3)\), radius \(\sqrt8\).
(a) Substitute \(y = x + k\): \(x^2 + (x + k)^2 - 4x - 6(x + k) + 5 = 0\), so \(x^2 + x^2 + 2kx + k^2 - 4x - 6x - 6k + 5 = 0\), giving \(2x^2 + (2k - 10)x + k^2 - 6k + 5 = 0\). ✓
(b) Tangent \(\Leftrightarrow\) discriminant = 0: \((2k - 10)^2 - 8(k^2 - 6k + 5) = 4k^2 - 40k + 100 - 8k^2 + 48k - 40 = -4k^2 + 8k + 60\).
\(-4(k^2 - 2k - 15) = -4(k - 5)(k + 3) = 0\), so \(k = 5\) or \(k = -3\).
\(k = 5\): \(2x^2 = 0\), contact at \((0, 5)\). \(k = -3\): \(2x^2 - 16x + 32 = 0\), so \((x - 4)^2 = 0\), contact at \((4, 1)\).
(c) Two points \(\Leftrightarrow\) discriminant \(> 0\): \(-4(k - 5)(k + 3) > 0 \Leftrightarrow (k - 5)(k + 3) < 0\), so \(-3 < k < 5\).
(d) \(k = 1\) gives \(2x^2 - 8x = 0\), so \(x = 0\) or 4: the points \((0, 1)\) and \((4, 5)\). Chord length \(= \sqrt{4^2 + 4^2} =\) \(4\sqrt2\), equal to the diameter \(2\sqrt8\), because \(y = x + 1\) passes through the centre.
Squaring \((x + k)\) as \(x^2 + k^2\), which loses the \(2kx\) term and spoils everything after it. Write the expansion out in full on its own line.
How do you find the circle through three points?
Any three points not on one straight line lie on exactly one circle. Two methods find it.
- Method 1, perpendicular bisectors: the centre is equidistant from all three points, so it lies on the perpendicular bisector of each chord. Find two of them and solve simultaneously; the radius is the distance from the centre to any of the points.
- Method 2, simultaneous equations: substitute each point into \(x^2 + y^2 + 2gx + 2fy + c = 0\) and solve the three linear equations for \(g\), \(f\) and \(c\).
- Shortcut: if two of the chords are perpendicular (gradient product \(-1\)), the third side is a diameter and its midpoint is the centre.
The points \(A(-1, 4)\), \(B(6, 5)\) and \(C(7, -2)\) lie on a circle.
(a) Find the equation of the perpendicular bisector of \(AB\). (b) The perpendicular bisector of \(BC\) has equation \(7y = x + 4\). Find the centre and radius of the circle. (c) Write down the equation of the circle.
(a) Midpoint of \(AB = (2.5, 4.5)\); gradient \(AB = \frac{5 - 4}{6 - (-1)} = \frac17\), so the perpendicular gradient is \(-7\). \(y - 4.5 = -7(x - 2.5)\), giving \(y = -7x + 22\).
(b) Solve with \(7y = x + 4\): \(7(-7x + 22) = x + 4\), so \(-49x + 154 = x + 4\), giving \(x = 3\), \(y = 1\): centre \((3, 1)\). Radius² \(= (-1 - 3)^2 + (4 - 1)^2 = 25\): radius 5.
(c) \((x - 3)^2 + (y - 1)^2 = 25\)
Check by method 2: the three points give \(-2g + 8f + c = -17\), \(12g + 10f + c = -61\) and \(14g - 4f + c = -53\), with solution \(g = -3\), \(f = -1\), \(c = -15\): centre \((3, 1)\), \(r^2 = 9 + 1 + 15 = 25\). ✓
Before you find any bisectors, check the gradients of the three sides: a product of \(-1\) means a right angle, a diameter, and a much shorter route to the centre.
How long is a tangent from a point outside the circle?
- From a point \(P\) outside a circle there are two tangents, touching the circle at \(A\) and \(B\). They have equal lengths: \(PA = PB\).
- The radius \(CA\) is perpendicular to the tangent, so triangle \(PAC\) has a right angle at \(A\): \(PA^2 = PC^2 - r^2\).
- Shortcut: if the circle is \(x^2 + y^2 + 2gx + 2fy + c = 0\), substituting \(P\) into the left-hand side gives \(PA^2\) directly. A positive value confirms \(P\) is outside.
- The quadrilateral \(PACB\) is a kite made of two congruent right-angled triangles: area \(= 2 \times \frac12 \times r \times PA = r \times PA\).
- To find the equations of the tangents from \(P\): take a line through \(P\) with gradient \(m\), substitute into the circle and set the discriminant to zero.
The circle \(C\) has equation \(x^2 + y^2 - 10x - 4y + 20 = 0\). The point \(P\) has coordinates \((1, -1)\).
(a) Find the centre and radius of \(C\). (b) Show that \(P\) lies outside \(C\), and find the length of a tangent from \(P\) to \(C\). (c) A line through \(P\) has gradient \(m\). Show that the \(x\)-coordinates of its points of intersection with \(C\) satisfy \((1 + m^2)x^2 - (2m^2 + 6m + 10)x + m^2 + 6m + 25 = 0\). (d) Hence find the equations of the two tangents from \(P\) to \(C\). (e) Find the area of the quadrilateral formed by \(P\), the two points of contact and the centre, and the angle between the two tangents.
(a) \((x - 5)^2 - 25 + (y - 2)^2 - 4 + 20 = 0\), so \((x - 5)^2 + (y - 2)^2 = 9\): centre \((5, 2)\), radius 3.
(b) \(PC^2 = (1 - 5)^2 + (-1 - 2)^2 = 25 > 9\), so \(P\) is outside. Tangent length \(= \sqrt{25 - 9} =\) 4. Shortcut check: \(1^2 + (-1)^2 - 10(1) - 4(-1) + 20 = 16 = 4^2\). ✓
(c) The line is \(y = m(x - 1) - 1\), so \(y - 2 = mx - (m + 3)\). Substitute into the centre–radius form: \((x - 5)^2 + (mx - (m + 3))^2 = 9\), so \(x^2 - 10x + 25 + m^2x^2 - 2m(m + 3)x + m^2 + 6m + 9 - 9 = 0\), giving \((1 + m^2)x^2 - (2m^2 + 6m + 10)x + m^2 + 6m + 25 = 0\). ✓
(d) Tangent \(\Leftrightarrow\) discriminant = 0. Take out the factor 4 first: \(4\left[(m^2 + 3m + 5)^2 - (1 + m^2)(m^2 + 6m + 25)\right] = 0\).
\((m^4 + 6m^3 + 19m^2 + 30m + 25) - (m^4 + 6m^3 + 26m^2 + 6m + 25) = -7m^2 + 24m = 0\), so \(m(24 - 7m) = 0\), giving \(m = 0\) or \(m = \frac{24}{7}\).
Tangents: \(y = -1\) and \(24x - 7y - 31 = 0\) (from \(y + 1 = \frac{24}{7}(x - 1)\)). Points of contact: \(m = 0\) gives \(x^2 - 10x + 25 = 0\), so \(A(5, -1)\); \(m = \frac{24}{7}\) gives \((25x - 53)^2 = 0\), so \(B\left(\frac{53}{25}, \frac{71}{25}\right)\).
(e) A kite of two right-angled triangles with legs 3 and 4: area \(= 2 \times \frac12 \times 3 \times 4 =\) 12. Half-angle at \(P\): \(\tan\theta = \frac34\), so the angle between the tangents \(= 2 \times 36.87° =\) 73.7°.
Forgetting that a vertical or horizontal tangent may be one of the answers. Here \(m = 0\) is a genuine root: dividing the discriminant equation by \(m\) would lose the tangent \(y = -1\). Factorise instead of dividing.
Exam tips for P2.2
A quick sketch with the centre, the radius and any given points shows which properties are in play (a right angle, a chord, a tangent) and catches impossible answers, such as a "tangent" that clearly cuts the circle.
"Exact radius" or "exact length" means \(\sqrt{12.5} = \frac{5\sqrt2}{2}\) or \(4\sqrt2\), not 3.54 or 5.66. Simplify the surd; give a decimal only when the question asks for one.
Name the property in words: "tangent ⊥ radius", "angle in a semicircle = 90°", "the perpendicular bisector of a chord passes through the centre". It turns a gradient calculation into a complete argument.
For "find the set of values", write the discriminant inequality, find the critical values, then decide inside or outside from a sketch of the quadratic in \(k\). Give the answer as an inequality such as \(-3 < k < 5\).
After finding an intersection, a point of contact or a centre, substitute it back into the equations it should satisfy. One line of arithmetic catches most sign errors.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| \((x + 3)^2 + (y - 4)^2 = 16\) has centre \((3, -4)\). | The bracket is \(x - a\), so \(x + 3\) means \(a = -3\). | Centre \((-3, 4)\), radius 4. |
| The number on the right-hand side is the radius. | The equation comes from Pythagoras, so the right side is a squared distance. | Radius = √(right-hand side). |
| Every equation \(x^2 + y^2 + 2gx + 2fy + c = 0\) is a circle. | After completing the square the right-hand side can be zero or negative. | A circle needs \(g^2 + f^2 - c > 0\). |
| The tangent at \(P\) has the same gradient as the radius \(CP\). | The tangent is perpendicular to the radius. | Tangent gradient \(= -1 \div\) (gradient of \(CP\)). |
| The length of the tangent from \(P\) is the distance \(PC\) to the centre. | \(PC\) is the hypotenuse of the right-angled triangle formed with the radius, so it is longer than the tangent. | \(PA^2 = PC^2 - r^2\). |
| The centre of a circle through three points is the average of the three points. | That is the centroid of the triangle, which is not in general equidistant from the vertices. | The centre is where the perpendicular bisectors of the chords meet. |
How this topic connects
P1 straight lines and quadratics
Gradients, midpoints, the distance between two points and the rule \(m_1 m_2 = -1\) are used in almost every circle question. Completing the square finds the centre; the discriminant decides how a line meets a circle; quadratic inequalities give the set of values of a constant.
P1 trigonometry
Arc length \(r\theta\) and sector area \(\frac12 r^2\theta\) combine with circle equations when a question asks for the arc between two points or the area of a segment cut off by a chord.
P4 parametric equations and implicit differentiation
A circle can be written parametrically as \(x = a + r\cos t\), \(y = b + r\sin t\), and tangents can be found by implicit differentiation; the result always agrees with "tangent ⊥ radius".
Key formulae
| Formula | Use | In the formulae list? |
|---|---|---|
| \((x - a)^2 + (y - b)^2 = r^2\) | circle with centre \((a, b)\) and radius \(r\) | No — learn it |
| \(x^2 + y^2 + 2gx + 2fy + c = 0\): centre \((-g, -f)\), \(r = \sqrt{g^2 + f^2 - c}\) | centre and radius from the expanded form | No — derive it |
| \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\); midpoint \(\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\) | radius, diameter, centre from a diameter | No — learn it |
| \(m_1 m_2 = -1\) | tangent ⊥ radius; angle in a semicircle; perpendicular bisectors | No — learn it |
| \(b^2 - 4ac > 0\), \(= 0\), \(< 0\) | line cuts, touches or misses a circle | No — learn it |
| \(PT^2 = PC^2 - r^2\) | length of a tangent from an outside point \(P\) | No — derive it |
Check your understanding
1. Find the centre and radius of the circle \(x^2 + y^2 - 8x + 2y - 8 = 0\).
\((x - 4)^2 - 16 + (y + 1)^2 - 1 - 8 = 0\), so \((x - 4)^2 + (y + 1)^2 = 25\). Centre \((4, -1)\), radius 5.
2. Explain why \(x^2 + y^2 + 4x - 6y + 20 = 0\) is not the equation of a circle.
Completing the square gives \((x + 2)^2 + (y - 3)^2 = 4 + 9 - 20 = -7\). A sum of squares cannot be negative, so no points satisfy it.
3. Does the point \((6, 1)\) lie inside, on or outside the circle \((x - 3)^2 + (y + 3)^2 = 25\)?
\((6 - 3)^2 + (1 + 3)^2 = 9 + 16 = 25\), equal to \(r^2\), so the point lies on the circle.
4. Find the equation of the tangent to \(x^2 + y^2 = 25\) at the point \((4, 3)\).
The radius from \((0, 0)\) has gradient \(\frac34\), so the tangent has gradient \(-\frac43\): \(y - 3 = -\frac43(x - 4)\). \(4x + 3y = 25\).
5. Find the length of a tangent from \((7, 1)\) to the circle \((x - 1)^2 + (y - 1)^2 = 11\).
Distance to the centre squared: \(6^2 + 0^2 = 36\). Tangent length \(= \sqrt{36 - 11} = \sqrt{25}\). The tangent length is 5.
6. Find the set of values of \(c\) for which the line \(y = c\) meets the circle \((x - 2)^2 + (y - 3)^2 = 8\) at two distinct points.
Substitute \(y = c\): \((x - 2)^2 = 8 - (c - 3)^2\). Two distinct points need \(8 - (c - 3)^2 > 0\), so \(|c - 3| < \sqrt8 = 2\sqrt2\). \(3 - 2\sqrt2 < c < 3 + 2\sqrt2\).
Specification checklist
- (P2 3.1) I can write down the equation of a circle from its centre and radius, and find the centre and radius from \((x - a)^2 + (y - b)^2 = r^2\).
- (P2 3.1) I can complete the square on \(x^2 + y^2 + 2gx + 2fy + c = 0\) (dividing through first if necessary) to find the centre and radius, and say when the equation is not a circle.
- (P2 3.1) I can use the facts that the angle in a semicircle is a right angle and that the perpendicular from the centre to a chord bisects the chord.
- (P2 3.1) I can find the equation of the tangent and the normal at a point on a circle, using the fact that the tangent is perpendicular to the radius.
- (P2 3.1) I can find where a line meets a circle and use the discriminant to decide whether the line cuts, touches or misses it, including lines with an unknown constant.
- (P2 3.1) I can find the equation of the circle through three given points.
- (P2 3.1) I can find the length of a tangent from a point outside a circle, and the equations of the tangents from that point.