Straight lines and circles – Year 12 A level maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. Gradient \(=\dfrac{\text{change in }y}{\text{change in }x}=\dfrac{y_2-y_1}{x_2-x_1}\). Gradient \(=\dfrac{7-(-1)}{8-2}=\dfrac86=\) \(\dfrac43\). The line rises 4 units for every 3 units across.
   2. Use \(y-y_1=m(x-x_1)\) with the point \(A(2,-1)\), then clear the fraction and collect the terms on one side. \(y+1=\tfrac43(x-2)\). Multiply by 3: \(3y+3=4x-8\), so \(4x-3y-11=0\). Check with \(B\): \(32-21-11=0\) ✓.
   3. The midpoint is the mean of the \(x\)-coordinates and the mean of the \(y\)-coordinates. \(M=\left(\dfrac{2+8}2,\dfrac{-1+7}2\right)=\) \((5,3)\).
   4. Use Pythagoras with \(\Delta x=6\) and \(\Delta y=8\). \(AB=\sqrt{6^2+8^2}=\sqrt{100}=\) 10. So \(AB\) is 10 units long.

2. Answer to Question 2

   1. Rearrange into the form \(y=mx+c\). \(4y=-3x+12\), so \(y=-\tfrac34x+3\) and the gradient is \(-\tfrac34\).
   2. Parallel lines have the same gradient. \(y+2=-\tfrac34(x-4)\), so \(4y+8=-3x+12\) and \(3x+4y=4\).
   3. Perpendicular gradients multiply to \(-1\), so the new gradient is \(-1\div\left(-\tfrac34\right)=\tfrac43\). The perpendicular gradient is \(\tfrac43\): \(y=\tfrac43x+5\), i.e. \(4x-3y+15=0\).
   4. The foot of the perpendicular is where the perpendicular line from part (c) meets \(L\); solve the two equations simultaneously. Solve \(3x+4y=12\) and \(4x-3y=-15\). Multiply: \(9x+12y=36\) and \(16x-12y=-60\). Add: \(25x=-24\), \(x=-\tfrac{24}{25}\), \(y=\tfrac{93}{25}\). Foot \((-\tfrac{24}{25},\tfrac{93}{25})\). Distance \(=\sqrt{\tfrac{576}{625}+\left(5-\tfrac{93}{25}\right)^2}=\sqrt{\tfrac{576+1024}{625}}=\sqrt{\tfrac{1600}{625}}=\) \(\tfrac85=1.6\). So the shortest distance from \((0,5)\) to \(L\) is 1.6 units.

3. Answer to Question 3

   1. A circle with centre \((a,b)\) and radius \(r\) has equation \((x-a)^2+(y-b)^2=r^2\). \((x-3)^2+(y+2)^2=25\). Note the signs: the centre \((3,-2)\) gives \((x-3)\) and \((y+2)\).
   2. Expand both brackets and collect all terms on one side. \(x^2-6x+9+y^2+4y+4=25\), so \(x^2+y^2-6x+4y-12=0\). This is the required form.
   3. Substitute each point into the left-hand side of the equation and compare with \(r^2=25\). \((6,2)\): \(3^2+4^2=25\) ✓. \((0,2)\): \((-3)^2+4^2=25\) ✓. Both points are on the circle.
   4. Compare the squared distance from the centre with \(r^2\): less means inside, more means outside. \((7,1)\): \((7-3)^2+(1+2)^2=16+9=25\). This equals \(r^2\), so the point lies on the circle.

4. Answer to Question 4

   1. Complete the square in \(x\) and in \(y\). Complete the square: \((x-4)^2-16+(y+3)^2-9-11=0\), so \((x-4)^2+(y+3)^2=36\). Centre \((4,-3)\), radius 6.
   2. Use the same method. \((x+1)^2-1+(y-2)^2-4-4=0\), so \((x+1)^2+(y-2)^2=9\). Centre \((-1,2)\), radius 3.
   3. Two circles meet at two points when the distance between the centres lies between the difference and the sum of the radii. Distance \(=\sqrt{5^2+5^2}=5\sqrt2\approx7.07\). The sum of radii is 9 and the difference is 3. Since \(3<7.07<9\) the circles intersect at two points.

5. Answer to Question 5

   1. The centre of the circle is the midpoint of the diameter, and the radius is half the length of the diameter. Centre = midpoint \(=(4,6)\). \(AB=\sqrt{36+64}=10\), so the radius is 5.
   2. \((x-4)^2+(y-6)^2=25\).
   3. Substitute \(C\) into the equation. \((8-4)^2+(9-6)^2=16+9=25\) ✓. So \(C\) lies on the circle.
   4. Gradient of \(AC=\dfrac{9-2}{8-1}=1\) and gradient of \(BC=\dfrac{9-10}{8-7}=-1\). The product is \(-1\) so \(AC\perp BC\). This illustrates the angle in a semicircle is a right angle theorem.

6. Answer to Question 6

   1. Substitute the point into the equation. \((4-1)^2+(2+2)^2=9+16=25\) ✓.
   2. The centre is \((1,-2)\). Centre \((1,-2)\): gradient \(=\dfrac{2-(-2)}{4-1}=\dfrac43\).
   3. A tangent is perpendicular to the radius at the point of contact. The tangent is perpendicular to the radius: gradient \(-\tfrac34\). \(y-2=-\tfrac34(x-4)\), \(4y-8=-3x+12\), so \(3x+4y=20\).
   4. Find both intercepts of the tangent. When \(y=0\): \(x=\tfrac{20}3\), so \(T(\tfrac{20}3,0)\). When \(x=0\): \(y=5\). Area \(=\tfrac12\times\tfrac{20}3\times5=\) \(\tfrac{50}3\approx16.7\) square units.

7. Answer to Question 7

   1. The mast is the centre and the range 5 km is the radius. \((x-3)^2+(y-4)^2=25\).
   2. Substitute the equation of the road into the equation of the circle. Substitute \(y=x-2\): \((x-3)^2+(x-6)^2=25\), so \(2x^2-18x+45=25\), giving \(2x^2-18x+20=0\), i.e. \(x^2-9x+10=0\). This is the required equation.
   3. Use the quadratic formula; the road has gradient 1, so the chord length is \(\sqrt2\) times the difference in \(x\). \(x=\dfrac{9\pm\sqrt{81-40}}2=\dfrac{9\pm\sqrt{41}}2\). The \(x\)-values differ by \(\sqrt{41}\) and \(y=x-2\) so the \(y\)-values also differ by \(\sqrt{41}\). Length \(=\sqrt{41+41}=\sqrt{82}=\) 9.06 km. About 9.06 km of road has coverage.
   4. The shortest distance is along the perpendicular from \(M\) to the road. The perpendicular from \(M\) has gradient \(-1\): \(y=-x+7\). It meets the road where \(x-2=-x+7\), \(x=4.5\), \(y=2.5\). Distance \(=\sqrt{1.5^2+1.5^2}=\sqrt{4.5}=\) 2.12 km. Since \(2.12<5\), this confirms the road passes through the coverage area.

8. Answer to Question 8

   1. Compare with \((x-a)^2+(y-b)^2=r^2\). Centre \((6,5)\), radius 3 m.
   2. From the path, \(y=\dfrac{53-3x}4\). Then \(y-5=\dfrac{33-3x}4=\dfrac{3(11-x)}4\). Substituting: \((x-6)^2+\dfrac9{16}(x-11)^2=9\). Multiply by 16: \(16x^2-192x+576+9x^2-198x+1089=144\), so \(25x^2-390x+1521=0\). The discriminant is \(390^2-4\times25\times1521=152100-152100=0\), so there is exactly one point of contact: the path is a tangent.
   3. \(x=\dfrac{390}{50}=7.8\) and \(y=\dfrac{53-23.4}4=7.4\). Contact point \((7.8,\,7.4)\). (Check: \(1.8^2+2.4^2=9\) ✓.)
   4. The other tangent is on the opposite side of the centre: the centre has \(3x+4y=18+20=38\), and \(53-38=15\), so the other is \(38-15=23\): \(3x+4y=23\).

9. Answer to Question 9

   1. The centre is at \((0,12)\) and the radius is 10. \(x^2+(y-12)^2=100\). Every cabin is 10 m from the centre of the wheel.
   2. Substitute \(x=6\) and solve for \(y\). \(x=6\): \((y-12)^2=64\), so \(y-12=\pm8\) and \(y=20\) or \(y=4\). Heights 20 m and 4 m. One position is above the centre and one below.
   3. The lowest point is directly below the centre, one radius down. Lowest point: \(12-10=\) 2 m above the ground. The lowest cabin is 2 m above the ground.
   4. Substitute \(x=5\) and take the positive square root, because the cabin is in the upper half. \((y-12)^2=75\), \(y=12+\sqrt{75}=12+8.66=\) 20.7 m. The cabin is about 20.7 m above the ground.

10. Answer to Question 10

    1. The perpendicular bisector of a chord passes through the centre of the circle. \(AB\) is horizontal with midpoint \((4,0)\), so its perpendicular bisector is \(x=4\). \(AC\) is vertical with midpoint \((0,3)\), so its perpendicular bisector is \(y=3\).
    2. They meet at the centre \((4,3)\). Radius \(=\sqrt{4^2+3^2}=\) 5 m.
    3. Expand the equation in centre-radius form. \((x-4)^2+(y-3)^2=25\Rightarrow\) \(x^2+y^2-8x-6y=0\). (Check \(B\): \(64-64=0\) ✓; \(C\): \(36-36=0\) ✓.)
    4. \(BC=\sqrt{64+36}=10=2\times5\), so \(BC\) is a diameter. The angle at \(A\) is \(90^\circ\) (angle in a semicircle).

11. Answer to Question 11

    1. Find the gradient of the road, then use \(y-y_1=m(x-x_1)\); the fence has the same gradient. Road gradient \(=\dfrac{5-2}{6-0}=\tfrac12\): \(y=\tfrac12x+2\). Fence: \(y-6=\tfrac12(x-2)\), so \(y=\tfrac12x+5\). The road is \(y=\tfrac12x+2\) and the fence is \(y=\tfrac12x+5\).
    2. A track at right angles to the road has gradient \(-1\div\tfrac12=-2\). Perpendicular gradient \(-2\): \(y-6=-2(x-2)\), so \(y=-2x+10\). The track is \(y=-2x+10\).
    3. Solve the equations of the track and the road simultaneously. \(\tfrac12x+2=-2x+10\) gives \(2.5x=8\), \(x=3.2\), \(y=3.6\). Point \((3.2,\,3.6)\). The track meets the road at \((3.2,\,3.6)\).
    4. The shortest distance between the parallel lines is the length of the perpendicular track. Distance \(=\sqrt{(3.2-2)^2+(3.6-6)^2}=\sqrt{1.44+5.76}=\sqrt{7.2}=\) 2.68 m. The fence is about 2.68 m from the road.

12. Answer to Question 12

    1. Two circles overlap when the distance between centres is between the difference and the sum of the radii. Centres \((2,3)\) and \((8,3)\): distance \(=6\). Radii 4 and 3: \(4-3=1<6<7=4+3\), so the circles overlap.
    2. Subtracting the equations removes the \(x^2\) and \(y^2\) terms. Subtract the equations: \((x-2)^2-(x-8)^2=16-9=7\). \((x^2-4x+4)-(x^2-16x+64)=12x-60=7\), so \(x=\dfrac{67}{12}=5.583\).
    3. Substitute \(x=\tfrac{67}{12}\) into the first circle. \((x-2)^2=\left(\tfrac{43}{12}\right)^2=12.84\), so \((y-3)^2=16-12.84=3.16\), \(y-3=\pm1.78\). So \(y=\) 4.78 or 1.22.
    4. The two intersection points have the same \(x\)-coordinate, so the chord is vertical. Chord length \(=2\times1.78=\) 3.56 m. The common chord is about 3.56 m long.

13. Answer to Question 13

    1. Substitute \(P\) into the equation of the circle. \((5-2)^2+(3-1)^2=9+4=13\) ✓. So \(P\) lies on the boundary.
    2. The tangent is perpendicular to the radius from the centre \((2,1)\) to \(P\). Radius gradient \(=\dfrac{3-1}{5-2}=\tfrac23\). Tangent gradient \(=-\tfrac32\). \(y-3=-\tfrac32(x-5)\), so \(2y-6=-3x+15\), giving \(3x+2y=21\). The tangent is \(3x+2y=21\).
    3. Set \(y=0\) and then \(x=0\) in the equation of the tangent. \(y=0\): \(x=7\). \(x=0\): \(y=10.5\). The ends are \((7,0)\) and \((0,10.5)\).
    4. Use Pythagoras with the two intercepts. Length \(=\sqrt{7^2+10.5^2}=\sqrt{159.25}=\) 12.62 km. The track is about 12.62 km long.

14. Answer to Question 14

    1. Use the midpoint formula and the gradient formula. Midpoint \((5,4)\). Gradient \(=\dfrac{6-2}{9-1}=\tfrac12\). The midpoint is \((5,4)\) and the gradient is \(\tfrac12\).
    2. Every point equidistant from \(A\) and \(B\) lies on the perpendicular bisector, which passes through the midpoint with gradient \(-2\). Perpendicular gradient \(-2\): \(y-4=-2(x-5)\), so \(y=-2x+14\). The perpendicular bisector is \(y=-2x+14\).
    3. The bus stop is on both the perpendicular bisector and the road, so solve simultaneously. \(x+2=-2x+14\) gives \(3x=12\), \(x=4\), \(y=6\). The bus stop is \((4,6)\).
    4. Use Pythagoras for each distance. To \(A\): \(\sqrt{3^2+4^2}=5\). To \(B\): \(\sqrt{5^2+0^2}=5\). ✓ The bus stop is 5 km from each village.

15. Answer to Question 15

    1. The perpendicular bisector of a chord passes through the centre. \(PQ\) is horizontal with midpoint \((4,1)\), so the bisector is \(x=4\).
    2. \(PR\) has gradient \(\dfrac{5-1}{8-2}=\tfrac23\) and midpoint \((5,3)\). The bisector has gradient \(-\tfrac32\): \(y-3=-\tfrac32(x-5)\).
    3. The centre lies on both bisectors. At \(x=4\): \(y=3+\tfrac32=4.5\). Centre \((4,4.5)\). Radius\(^2=(2-4)^2+(1-4.5)^2=4+12.25=16.25\), so the radius is \(\sqrt{16.25}=\dfrac{\sqrt{65}}2\approx4.03\) m.
    4. Substitute \(R\) into \((x-4)^2+(y-4.5)^2=16.25\). \(R\): \((8-4)^2+(5-4.5)^2=16+0.25=16.25\) ✓. Area \(=\pi r^2=16.25\pi=\) 51.1 m\(^2\).

16. Answer to Question 16

    1. Substitute the line into the circle and expand \((2x+k)^2=4x^2+4kx+k^2\). Substitute: \(x^2+(2x+k)^2=20\Rightarrow5x^2+4kx+k^2-20=0\).
    2. A tangent gives equal roots, so the discriminant is zero. Discriminant \(=16k^2-20(k^2-20)=400-4k^2=0\), so \(k^2=100\) and \(k=\pm10\). (For \(k=10\) the contact is at \((-4,2)\).)
    3. Two points need \(400-4k^2>0\): \(-10\lt k<10\).
    4. \(5x^2+20x+5=0\Rightarrow x^2+4x+1=0\Rightarrow x=-2\pm\sqrt3\). The points are \((-2+\sqrt3,\,1+2\sqrt3)\) and \((-2-\sqrt3,\,1-2\sqrt3)\). The \(x\)-difference is \(2\sqrt3\) and the \(y\)-difference is \(4\sqrt3\), so the chord is \(\sqrt{12+48}=\sqrt{60}=2\sqrt{15}\approx7.75\).

17. Answer to Question 17

    1. Complete the square in \(x\) and in \(y\). \((x-3)^2+(y-2)^2=25\). Centre \((3,2)\), radius 5.
    2. \(36+36-36-24-12=0\) ✓. Radius gradient \(\dfrac{6-2}{6-3}=\tfrac43\); tangent gradient \(-\tfrac34\): \(y-6=-\tfrac34(x-6)\), so \(3x+4y=42\).
    3. \(y=0\): \(x=14\). \(T=\) \((14,0)\).
    4. \(TP=\sqrt{(14-6)^2+(0-6)^2}=\sqrt{64+36}=10\) ✓. The distance from \(T\) to the centre is \(\sqrt{11^2+2^2}=\sqrt{125}\). By Pythagoras the tangent length is \(\sqrt{125-25}=10\) for either tangent from \(T\), so the other tangent is also 10.

18. Answer to Question 18

    1. Centres \((0,0)\) and \((6,0)\), each with radius 5.
    2. Subtract: \(x^2-(x-6)^2=0\Rightarrow12x-36=0\), so \(x=3\). Then \(y^2=25-9=16\), \(y=\pm4\). Points \((3,4)\) and \((3,-4)\).
    3. The chord from \((3,-4)\) to \((3,4)\) has length 8 km. It is vertical, and the line of centres is horizontal, so they are perpendicular.
    4. The midpoint of the centres is \((3,0)\), which lies on the line \(x=3\). ✓

19. Answer to Question 19

    1. Substitute each point into \(x^2+y^2\). \(3^2+4^2=25\), \((-4)^2+3^2=25\), \(0+(-5)^2=25\) ✓.
    2. \(P\): \(AP=\tfrac{4}{8}=\tfrac12\), \(BP=\tfrac{4}{-2}=-2\); product \(-1\). \(Q\): \(AQ=\tfrac3{1}=3\), \(BQ=\tfrac{3}{-9}=-\tfrac13\); product \(-1\). \(R\): \(AR=\tfrac{-5}{5}=-1\), \(BR=\tfrac{-5}{-5}=1\); product \(-1\).
    3. Gradient \(TA=\dfrac y{x+5}\) and gradient \(TB=\dfrac y{x-5}\). Their product is \(\dfrac{y^2}{x^2-25}\). Since \(x^2+y^2=25\), \(x^2-25=-y^2\), so the product is \(\dfrac{y^2}{-y^2}=-1\) (for \(y\ne0\)). Hence \(TA\perp TB\) for every point on the circle: the angle in a semicircle is a right angle.

20. Answer to Question 20

    1. The circle touches the \(x\)-axis, so the distance from the centre to the axis equals the radius: the centre is at height \(a\), so \(r=a\). Equation: \((x-a)^2+(y-a)^2=a^2\).
    2. Substitute \((8,4)\): \((8-a)^2+(4-a)^2=a^2\Rightarrow64-16a+a^2+16-8a+a^2=a^2\Rightarrow a^2-24a+80=0\). \((a-4)(a-20)=0\), so \(a=4\) or \(20\). Circles: \((x-4)^2+(y-4)^2=16\) and \((x-20)^2+(y-20)^2=400\).
    3. Centres \((4,4)\) and \((20,20)\): distance \(=16\sqrt2\approx22.6\). Radii 4 and 20 have sum 24 and difference 16. Since \(16<22.6<24\) the circles intersect at two points. Both circles are symmetrical about the line \(y=x\) (their centres lie on it), so since \((8,4)\) is an intersection point, its reflection \((4,8)\) is the other. Check: \((4-4)^2+(8-4)^2=16\) ✓ and \((4-20)^2+(8-20)^2=256+144=400\) ✓.