9709 Pure 1 · 1.3 Coordinate Geometry · Session 4 of 4

Chords and the Semicircle

Two facts about circles that Cambridge uses again and again: where the center sits compared with any chord, and the right angle hiding in every semicircle. Then we finish two real exam questions and try a third that joins the whole topic together.

Do Now, and why this matters7 min

The chord fact, and a 7-mark Cambridge question16 min

The angle in a semicircle, both ways round10 min

Finishing June 2023 Q12: parts (c) and (d)9 min

Review: one question that uses all of 1.39 min

Recap and exit ticket7 min

Do Now · 5 min · no notes, no calculator

\(A\) is \((-1, 7)\) and \(B\) is \((6, 0)\). Find the perpendicular bisector of \(AB\). Does it pass through \((2, 3)\)?

You did this in Session 1: midpoint, gradient, flip and change sign, then the line. Write every step, then tap to check.

Tap to check.

midpoint: \(\left(\dfrac{-1 + 6}{2}, \dfrac{7 + 0}{2}\right) = \left(\tfrac{5}{2}, \tfrac{7}{2}\right)\)

gradient \(AB\): \(\dfrac{0 - 7}{6 - (-1)} = \dfrac{-7}{7} = -1\) so the perpendicular gradient is \(1\)

line: \(y - \tfrac{7}{2} = 1\left(x - \tfrac{5}{2}\right)\), so \(y = x + 1\)

test \((2, 3)\): \(2 + 1 = 3\). Yes, it is on the line.

\(y = x + 1\), and \((2, 3)\) is on it.

chord \(AB\) \(y = x + 1\)

\((2, 3)\) is the center of a circle through \(A\) and \(B\): both are 5 away from it. That is no accident, and it is today's first fact.

Why this matters · 2 min

Find the center of a circle when all you have is its edge.

Most real circles do not come with their center marked. Today's two facts find it from points on the edge.

What each fact answers

The chord fact: pick any two points on the edge. The center is somewhere on their perpendicular bisector.

The angle in a semicircle: spot a right angle with its corner on the circle, and you have found a diameter.

Where you will see it

Sugar mill ruins. The old windmill towers across St. Croix are round stone walls. Measure three points on a wall and the chord fact puts the center on a map.

A carpenter's square. Set the corner of a square on the rim of a round tabletop and mark where its edges cross the rim. That line is a diameter. Do it twice: the crossing is the center.

Hubs. A Ferris wheel, a satellite dish or a bicycle wheel has to turn about its exact center, or it wobbles.

Direct instruction · 1 of 6 · the chord fact

The center of a circle lies on the perpendicular bisector of every chord.

Why. A chord joins two points on a circle (a diameter is the chord through the center). \(C\) is the center, so \(CA = r\) and \(CB = r\). That makes \(C\) the same distance from \(A\) as from \(B\). In Session 1 you saw that the points the same distance from \(A\) and \(B\) are exactly the points on the perpendicular bisector of \(AB\). So \(C\) is on it.

Two ways Cambridge uses it. (1) The line from the center to the midpoint \(M\) of a chord is perpendicular to the chord, so their gradients multiply to \(-1\). (2) Draw the perpendicular bisectors of two different chords: they cross at the center.

Direct instruction · 2 of 6 · Specimen Paper 1, Q10(b) · 4 marks

The circle \(x^2 + y^2 + 4x - 2y - 20 = 0\) has center \(C\) and passes through points \(A\) and \(B\).

It is given that the midpoint, \(D\), of \(AB\) has coordinates \(\left(1\tfrac{1}{2}, 1\tfrac{1}{2}\right)\). Find the equation of \(AB\), giving your answer in the form \(y = mx + c\). [4]

circle \(CD\) chord \(AB\)

Part (a), in Session 2: \((x + 2)^2 + (y - 1)^2 = 25\), so \(C\) is \((-2, 1)\) and \(r = 5\).

The plan: we know a point on \(AB\) (that is \(D\)) but not its gradient. The chord fact gives it: \(CD\) is perpendicular to \(AB\).

B1 · gradient \(CD\): \(\dfrac{1\tfrac{1}{2} - 1}{1\tfrac{1}{2} - (-2)} = \dfrac{\tfrac{1}{2}}{3\tfrac{1}{2}} = \dfrac{1}{7}\)

M1 · \(CD \perp AB\), so flip and change sign: gradient \(AB = -7\)

M1 · the line through \(D\) with gradient \(-7\): \(y - 1\tfrac{1}{2} = -7\left(x - 1\tfrac{1}{2}\right)\)

expand: \(y - 1\tfrac{1}{2} = -7x + 10\tfrac{1}{2}\) add \(1\tfrac{1}{2}\):

A1 · \(y = -7x + 12\)

From the examiners: if you start from \(y = mx + c\) instead, the method mark only comes once you have found \(c\). Either way, finish the line.

Direct instruction · 3 of 6 · Specimen Paper 1, Q10(c) · 3 marks

Find, by calculation, the \(x\)-coordinates of \(A\) and \(B\).

Why substitute: \(A\) and \(B\) are on the line and on the circle, so their coordinates fit both equations. This is Session 3's move.

M1 · put \(y = -7x + 12\) into the circle:
\(x^2 + (-7x + 12)^2 + 4x - 2(-7x + 12) - 20 = 0\)

expand each piece:
\((-7x + 12)^2 = 49x^2 - 168x + 144\) and \(-2(-7x + 12) = 14x - 24\)

collect: \(x^2 + 49x^2 - 168x + 144 + 4x + 14x - 24 - 20 = 0\)

A1: \(50x^2 - 150x + 100 = 0\)

divide by 50 and factor: \(x^2 - 3x + 2 = 0\), \((x - 1)(x - 2) = 0\)

A1 · \(x = 1\) or \(x = 2\)

the \(y\) values, from the line: \(-7(1) + 12 = 5\) and \(-7(2) + 12 = -2\), so \((1, 5)\) and \((2, -2)\)

check the midpoint: \(\left(\tfrac{1 + 2}{2}, \tfrac{5 + (-2)}{2}\right) = \left(1\tfrac{1}{2}, 1\tfrac{1}{2}\right)\) = \(D\) ✓

Show the method. The mark scheme wants the factoring (or the formula with numbers in) on the page. A calculator answer alone scores nothing for the last mark.

Direct instruction · 4 of 6 · semicircle

Join the ends of a diameter to any other point on the circle: the angle there is \(90^\circ\).

two ways to check it, with \(P(3, 4)\) on \(x^2 + y^2 = 25\), \(A(-5, 0)\), \(B(5, 0)\):
gradients: \(\tfrac{4 - 0}{3 - (-5)} = \tfrac{1}{2}\) and \(\tfrac{4 - 0}{3 - 5} = -2\); \(\tfrac{1}{2} \times (-2) = -1\), so \(AP \perp BP\)

Pythagoras: \(AP^2 = 8^2 + 4^2 = 80\), \(BP^2 = 2^2 + 4^2 = 20\)
\(80 + 20 = 100 = 10^2 = AB^2\), so the angle at \(P\) is a right angle

Direct instruction · 5 of 6 · the Cambridge move: use it backwards

A right angle with its corner on a circle means the other two points are the ends of a diameter.

\(P(-3, 1)\), \(Q(1, 5)\) and \(R(5, 1)\) lie on a circle. Show that \(PR\) is a diameter, and find the equation of the circle.

\(PQ\), \(QR\) diameter \(PR\)

gradient \(PQ\): \(\dfrac{5 - 1}{1 - (-3)} = \dfrac{4}{4} = 1\) gradient \(QR\): \(\dfrac{1 - 5}{5 - 1} = \dfrac{-4}{4} = -1\)

product \(1 \times (-1) = -1\), so \(PQ \perp QR\) and angle \(PQR = 90^\circ\)

Which pair is the diameter? The two points that are not the right-angle corner. The corner is at \(Q\), so the diameter is \(PR\) (the hypotenuse).

center = midpoint of \(PR\): \(\left(\tfrac{-3 + 5}{2}, \tfrac{1 + 1}{2}\right) = (1, 1)\)

radius = half of \(PR\): \(PR = 5 - (-3) = 8\), so \(r = 4\) check: \(Q\) is \(\sqrt{0^2 + 4^2} = 4\) from \((1, 1)\) ✓

\((x - 1)^2 + (y - 1)^2 = 16\)

Direct instruction · 6 of 6 · June 2023, Paper 11, Q12(c) · 3 marks

Find the equation of circle \(Q\) and hence verify that the \(y\)-coordinates of both points where the two circles meet are 11.

circle \(P\) tangent at \(A\) circle \(Q\)

From Session 2: circle \(P\) is \(x^2 + (y - 2)^2 = 100\) and the tangent at \(A(6, 10)\) is \(y = -\tfrac{3}{4}x + \tfrac{29}{2}\). Circle \(Q\) has its center where that tangent meets the \(y\)-axis, and radius \(\tfrac{5}{2}\sqrt{5}\).

M1 · on the \(y\)-axis \(x = 0\), so put \(x = 0\) in the tangent: \(y = \tfrac{29}{2}\), center \(\left(0, \tfrac{29}{2}\right)\)

radius squared: \(\left(\tfrac{5}{2}\sqrt{5}\right)^2 = \tfrac{25}{4} \times 5 = \tfrac{125}{4}\)

A1 · \(x^2 + \left(y - \tfrac{29}{2}\right)^2 = \tfrac{125}{4}\)

B1 · verify: put \(y = 11\) into both circles
\(P\): \(x^2 + 9^2 = 100\), so \(x^2 = 19\)
\(Q\): \(x^2 + \left(11 - \tfrac{29}{2}\right)^2 = \tfrac{125}{4}\), \(x^2 + \tfrac{49}{4} = \tfrac{125}{4}\), so \(x^2 = 19\)

Why that verifies it: with \(y = 11\), both circles give the same two points, \(\left(\pm\sqrt{19}, 11\right)\). So those points are on both circles.

What the examiners saw: many did not use part (b) to get the center, and most skipped the verify. The quick route is the one above: \(y = 11\) into both.

June 2023, Paper 11, Q12(d) · 3 marks

Find the coordinates of the points where the tangent meets circle \(Q\), giving the answers in surd form.

Surd form means exact, with the square root left in and simplified. \(\sqrt{20}\) simplifies because \(20 = 4 \times 5\): \(\sqrt{20} = \sqrt{4}\sqrt{5} = 2\sqrt{5}\). A decimal like 4.47 is not surd form.

circle \(Q\) the tangent from part (b)

Notice first: the tangent to \(P\) goes through the center of \(Q\). A line through the center of a circle cuts it twice, at the two ends of a diameter. So expect two points, the same distance either side of \(\left(0, \tfrac{29}{2}\right)\).

M1 · substitute \(y = -\tfrac{3}{4}x + \tfrac{29}{2}\), so \(y - \tfrac{29}{2} = -\tfrac{3}{4}x\):
\(x^2 + \left(-\tfrac{3}{4}x\right)^2 = \tfrac{125}{4}\)

square and add: \(x^2 + \tfrac{9}{16}x^2 = \tfrac{25}{16}x^2\), so \(\tfrac{25}{16}x^2 = \tfrac{125}{4}\)

multiply both sides by \(\tfrac{16}{25}\): \(x^2 = \tfrac{125 \times 16}{4 \times 25} = 20\)

A1 · square root, both signs: \(x = \pm\sqrt{20} = \pm 2\sqrt{5}\)

\(y\) from the line: \(x = 2\sqrt{5}\): \(y = \tfrac{29}{2} - \tfrac{3}{4}(2\sqrt{5}) = \tfrac{29}{2} - \tfrac{3}{2}\sqrt{5}\)
\(x = -2\sqrt{5}\): \(y = \tfrac{29}{2} + \tfrac{3}{2}\sqrt{5}\)

A1 · \(\left(2\sqrt{5}, \tfrac{29}{2} - \tfrac{3}{2}\sqrt{5}\right)\) and \(\left(-2\sqrt{5}, \tfrac{29}{2} + \tfrac{3}{2}\sqrt{5}\right)\)

Check: the midpoint of the two points is \(\left(0, \tfrac{29}{2}\right)\), the center of \(Q\), as expected for a diameter.

Review · you try first · June 2024, Paper 12, Q7(a) · 5 marks

The equation of a circle is \((x - 6)^2 + (y + a)^2 = 18\). The line \(y = 2a - x\) is a tangent to the circle.

(a) Find the two possible values of the constant \(a\). [5]

3 minutes on your own. This one needs Session 2 (reading a circle) and Session 3 (tangent means \(b^2 - 4ac = 0\)).

What the examiners saw: the substitute-then-discriminant route worked best. Marks were lost expanding brackets, and some forgot the 18 on the right-hand side.

M1 · replace \(y\) with \(2a - x\): \((x - 6)^2 + (2a - x + a)^2 = 18\), so \((x - 6)^2 + (3a - x)^2 = 18\)

expand: \(x^2 - 12x + 36 + 9a^2 - 6ax + x^2 = 18\)

A1 · everything on one side, take away 18: \(2x^2 - (12 + 6a)x + 9a^2 + 18 = 0\)
so \(A = 2\), \(B = -(12 + 6a)\), \(C = 9a^2 + 18\)

DM1 · tangent, so \(B^2 - 4AC = 0\): \((12 + 6a)^2 - 4(2)(9a^2 + 18) = 0\)

A1 · expand: \(144 + 144a + 36a^2 - 72a^2 - 144 = 0\), so \(-36a^2 + 144a = 0\)

A1 · \(-36a(a - 4) = 0\), \(a = 0\) or \(a = 4\)

Do both survive? The question asks for two values and puts no condition on \(a\), so yes. Put each back: \(a = 4\) gives \(2x^2 - 36x + 162 = 0\), \((x - 9)^2 = 0\), touching at \((9, -1)\); \(a = 0\) touches at \((3, -3)\).

Review · June 2024, Paper 12, Q7(b) · 3 marks

For the greater value of \(a\), find the equation of the diameter which is perpendicular to the given tangent.

circle, \(a = 4\) tangent \(y = 8 - x\) the diameter

What the examiners saw: about 30% left this part out, and many spent time finding the touching point when the center was all they needed. Draw the picture first.

Read the picture: a diameter goes through the center. Perpendicular to the tangent means it is the radius to the touching point, carried on across (tangent ⊥ radius, Session 2).

B1 · greater value \(a = 4\), so the circle is \((x - 6)^2 + (y + 4)^2 = 18\): center \((6, -4)\)

B1 · tangent \(y = 8 - x\) has gradient \(-1\), so the perpendicular gradient is \(1\)

line through \((6, -4)\), gradient 1: \(y + 4 = 1(x - 6)\)

B1 · \(y = x - 10\)

Check: the touching point \((9, -1)\) from part (a) fits: \(9 - 10 = -1\) ✓

Recap and homework · 3 min

Two facts, and a checklist for Test 1.

Chord fact: the center is on the perpendicular bisector of every chord. Center to midpoint is perpendicular to the chord: gradients multiply to \(-1\).

Angle in a semicircle: a diameter makes a \(90^\circ\) angle at any other point on the circle. Backwards: a right angle on the circle means the hypotenuse is a diameter, so its midpoint is the center.

1.3 checklist for Test 1 (Thursday, November 5; review day Tuesday, November 3): gradient, distance, midpoint · parallel and perpendicular lines · equation of a line · perpendicular bisector · center and radius from either form of a circle · inside, on or outside · tangent at a point · line meets curve with \(b^2 - 4ac\) · finding \(k\) for a tangent and where it touches · chord fact · angle in a semicircle.

Homework, as posted in Schoology. Workbook practice, pages 17 to 27, due Friday, October 30. Read section 3.10 of the textbook (The circle) and try Exercise 3F questions 11 and 18. Stuck? Write down where you got stuck and bring it; that still counts as done.

Exit ticket · 5 min · Schoology, Paper 1 → 1.3 Coordinate Geometry

Four questions. Work on paper first, then enter your answers in Schoology.

1. A circle has center \((4, -1)\). The midpoint of the chord \(AB\) is \(M(6, 2)\). What is the gradient of \(AB\)?

2. \(P(-2, 3)\), \(Q(2, 5)\) and \(R(4, 1)\) lie on a circle, and angle \(PQR = 90^\circ\). What is the center of the circle?

3. \(A(1, 4)\) and \(B(7, 2)\) lie on a circle. The center of the circle lies on which line?

4. The circle \(x^2 + y^2 + 2x - 8y - 8 = 0\) has center \(C\). The chord \(AB\) has midpoint \(M(1, 5)\). Find the equation of \(AB\) in the form \(y = mx + c\). Show your working.

Submit before you leave. An attempt left open scores nothing.

If time · the sugar mill problem

Three points on a round wall: \(A(2, 6)\), \(B(-4, 6)\), \(C(-5, -1)\). Find the center and the equation of the circle.

chords \(AB\), \(AC\) their perpendicular bisectors

Use the chord fact twice: the center is on both perpendicular bisectors. Tap to check.

\(AB\) is horizontal (both \(y = 6\)), midpoint \((-1, 6)\): its perpendicular bisector is the vertical line \(x = -1\)

\(AC\): midpoint \(\left(-\tfrac{3}{2}, \tfrac{5}{2}\right)\), gradient \(\tfrac{-1 - 6}{-5 - 2} = 1\), so perpendicular gradient \(-1\): \(y - \tfrac{5}{2} = -\left(x + \tfrac{3}{2}\right)\), \(y = -x + 1\)

where they cross, put \(x = -1\): \(y = 2\), center \((-1, 2)\)

radius, center to \(A\): \(\sqrt{3^2 + 4^2} = 5\)

\((x + 1)^2 + (y - 2)^2 = 25\)