Equation of a circle and tangents – Year 11 GCSE maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. \(x^2+y^2=25\). With centre \((0,0)\) the brackets become \(x^2\) and \(y^2\), and the right-hand side is \(r^2=5^2=25\).
   2. \((x-2)^2+(y-3)^2=16\). Substitute \(a=2\), \(b=3\) and \(r^2=4^2=16\) into \((x-a)^2+(y-b)^2=r^2\).
   3. \((x+1)^2+(y-4)^2=9\) (subtracting \(-1\) gives \(+1\)). Substitute \(a=-1\), \(b=4\) and \(r^2=9\).
   4. \(x^2+y^2=20\). Squaring \(\sqrt{20}\) gives 20, so the right-hand side is 20.
   5. \(r^2=(6-3)^2+(-1+5)^2=9+16=25\), so \((x-3)^2+(y+5)^2=25\). The radius is the distance from the centre to the given point, so we find \(r^2\) with Pythagoras.

2. Answer to Question 2

   1. Centre \((0,0)\), radius 7. Compare with \(x^2+y^2=r^2\): \(r^2=49\), so \(r=7\).
   2. Centre \((3,-2)\), radius 5. The signs inside the brackets are the opposite of the coordinates of the centre, and \(r=\sqrt{25}\).
   3. Centre \((-4,0)\), radius \(\sqrt{12}=\) \(2\sqrt3\). \(x+4=x-(-4)\) and \(y^2=(y-0)^2\). Simplify \(\sqrt{12}=\sqrt{4\times3}\).
   4. Centre \((1,1)\), radius \(\sqrt{18}=\) \(3\sqrt2\). Simplify \(\sqrt{18}=\sqrt{9\times2}\).

3. Answer to Question 3

   1. \((3,4)\): \(9+16=25\). \((5,1)\): \(25+1=26\). \((2,-3)\): \(4+9=13\). \((-4,-3)\): \(16+9=25\). \((4,4)\): \(16+16=32\).
   2. On: \((3,4)\) and \((-4,-3)\). Outside: \((5,1)\) and \((4,4)\) (greater than 25). Inside: \((2,-3)\) (less than 25).
   3. \((2,-3)\) has the smallest value, 13, so it is closest to the centre. So \((2,-3)\) is the closest point, at a distance \(\sqrt{13}\approx3.61\) from the centre.

4. Answer to Question 4

   1. Gradient of \(OA=\dfrac{4-0}{3-0}=\) \(\dfrac43\). Use gradient \(=\dfrac{\text{change in }y}{\text{change in }x}\) from \(O(0,0)\) to \(A(3,4)\).
   2. Perpendicular gradient \(=-1\div\dfrac43=\) \(-\dfrac34\). Perpendicular gradients multiply to \(-1\), so the tangent gradient is the negative reciprocal of \(\tfrac43\).
   3. \(y-4=-\dfrac34(x-3)\). Multiply by 4: \(4y-16=-3x+9\), so \(3x+4y=25\). Check: \(3\times3+4\times4=25\). Use \(y-y_1=m(x-x_1)\) with the point \(A\), then clear the fraction.

5. Answer to Question 5

   1. \((-6)^2+8^2=36+64=100\). The point satisfies the equation, so \(P\) is on the circle.
   2. Gradient of \(OP=\dfrac{8}{-6}=-\dfrac43\). The tangent gradient is the negative reciprocal: \(\dfrac34\). The tangent is perpendicular to the radius \(OP\).
   3. \(y-8=\dfrac34(x+6)\), so \(y=\dfrac34x+4.5+8=\) \(y=\dfrac34x+12.5\). Check: when \(x=-6\), \(y=-4.5+12.5=8\). Use \(y-y_1=m(x-x_1)\) with the point \(P(-6,8)\), then expand and add 8.

6. Answer to Question 6

   1. \((5,0)\), \((-5,0)\), \((0,5)\), \((0,-5)\). The circle has radius 5 and centre \(O\), so it meets each axis 5 units from the origin.
   2. The radius to \((5,0)\) is horizontal, so the tangent is vertical: \(x=5\). A tangent is perpendicular to the radius at the point of contact, and every point on a vertical line has the same \(x\)-coordinate.
   3. The radius to \((0,-5)\) is vertical, so the tangent is horizontal: \(y=-5\). Every point on a horizontal line has the same \(y\)-coordinate.
   4. The tangents \(x=\pm5\) and \(y=\pm5\) form a square of side 10, area 100. Each side of the square runs from \(-5\) to 5, so the side is 10 and the area is \(10^2=100\) square units.

7. Answer to Question 7

   1. \(x^2+(y-10)^2=100\). The centre is \((0,10)\) and the radius is 10, so \(r^2=100\).
   2. \(6^2+(y-10)^2=100\), so \((y-10)^2=64\) and \(y-10=\pm8\). The heights are \(y=18\) m and \(y=2\) m. Substitute \(x=6\), subtract 36 from both sides and take the square root, remembering \(\pm\). The pod is either 18 m or 2 m above the ground.
   3. \(6^2+(2-10)^2=36+64=100\), so \((6,2)\) is on the circle. The point satisfies the equation, so the pod is on the wheel.
   4. The lowest point is \((0,0)\) (\(y-10=-10\)): it touches the ground (0 m). The lowest point is directly below the centre, one radius down: \(10-10=0\). So the wheel just touches the ground.

8. Answer to Question 8

   1. \(x^2+y^2=50^2=\) \(x^2+y^2=2500\). The radar range is a circle with centre \(O\) and radius 50 km.
   2. \(30^2+40^2=900+1600=2500\), so it is on the boundary. Compare \(x^2+y^2\) with \(r^2=2500\): equal means on the circle.
   3. \(35^2+40^2=2825>2500\), so it is outside: not detected. Its distance is \(\sqrt{2825}=53.15\), which is \(53.15-50=\) 3.15 km beyond the edge. A value bigger than 2500 means the ship is further than 50 km from \(O\).
   4. \(20^2+30^2=1300\lt 2500\), so it is inside and detected. \(\sqrt{1300}\approx36.1\) km, which is less than 50 km.

9. Answer to Question 9

   1. \((x-3)^2+(y+2)^2=36\). The centre is \((3,-2)\), so the brackets are \((x-3)\) and \((y+2)\), and \(r^2=6^2=36\).
   2. \((3-3)^2+(4+2)^2=0+36=36\). ✓ The point satisfies the equation, so it is on the boundary.
   3. Put \(y=-2\): \((x-3)^2=36\), so \(x=9\) or \(x=-3\). The point on the right is \((9,-2)\).
   4. \((6,2)\): \((3)^2+(4)^2=25\lt 36\), so it is inside (watered). \((8,3)\): \((5)^2+(5)^2=50>36\), so it is outside (not watered).

10. Answer to Question 10

    1. \(24^2+32^2=576+1024=1600=40^2\). The point satisfies \(x^2+y^2=40^2\), so \(A\) is on the track.
    2. Gradient of \(OA=\dfrac{32}{24}=\dfrac43\); tangent gradient \(-\dfrac34\). \(y-32=-\dfrac34(x-24)\), so \(4y-128=-3x+72\) and \(3x+4y=200\). The tangent is perpendicular to the radius \(OA\). Use \(y-y_1=m(x-x_1)\), then multiply by 4. Check: \(72+128=200\).
    3. \(y=0\): \(x=\dfrac{200}3=66.7\). \(x=0\): \(y=50\). So \((66.7,0)\) and \((0,50)\). Put \(y=0\) to find where the tangent meets the \(x\)-axis, and \(x=0\) for the \(y\)-axis.
    4. From \((24,32)\) to \((0,50)\): \(\sqrt{24^2+18^2}=\sqrt{900}=\) 30 m. Use Pythagoras with the changes \(24\) across and \(50-32=18\) up. The runner travels 30 m.

11. Answer to Question 11

    1. Substitute \(y=x+1\): \(x^2+(x+1)^2=25\), so \(2x^2+2x+1=25\), \(2x^2+2x-24=0\), so \(x^2+x-12=0\). Replace \(y\) by \(x+1\), expand \((x+1)^2=x^2+2x+1\), subtract 25 and divide by 2.
    2. \((x+4)(x-3)=0\), so \(x=3\) or \(x=-4\). The points are \((3,4)\) and \((-4,-3)\). We need two numbers with product \(-12\) and sum 1: 4 and \(-3\). Find each \(y\) from \(y=x+1\).
    3. Distance \(=\sqrt{(3-(-4))^2+(4-(-3))^2}=\sqrt{49+49}=\sqrt{98}=\) 9.90 m. The changes are 7 across and 7 up, so we use Pythagoras. About 9.90 m of the path is inside the pond.

12. Answer to Question 12

    1. \(x^2+(x+4)^2=8\), so \(2x^2+8x+16=8\), so \(2x^2+8x+8=0\), so \(x^2+4x+4=0\).
    2. \((x+2)^2=0\), so \(x=-2\) is a repeated root: one point of contact only. A line meeting a circle at exactly one point is a tangent.
    3. \(y=-2+4=2\), so the point of contact is \((-2,2)\).
    4. Gradient of the radius \(=\dfrac{2}{-2}=-1\). Gradient of the line \(=1\). \(-1\times1=-1\), so they are perpendicular.

13. Answer to Question 13

    1. The radius \(OA\) is perpendicular to the tangent \(TA\). A tangent is always perpendicular to the radius at the point of contact.
    2. \(OT=13\) and \(OA=5\). \(TA^2=13^2-5^2=144\), so \(TA=\) 12 m. \(OT=13\) is the hypotenuse because it is opposite the right angle.
    3. \(\sin(ATO)=\dfrac5{13}\), so \(ATO=\) \(22.6^\circ\). \(OA=5\) is opposite angle \(ATO\) and \(OT=13\) is the hypotenuse, so we use sine.
    4. Two congruent triangles: \(2\times\tfrac12\times5\times12=\) 60 m². The kite is made of the two right-angled triangles \(OAT\) and \(OBT\), each with base 5 and height 12.

14. Answer to Question 14

    1. Angle \(AOB=90^\circ\) because the axes are perpendicular, and the angle in a semicircle is \(90^\circ\) (converse), so \(AB\) is a diameter.
    2. The centre is the midpoint of \(AB\): \((4,3)\). \(AB=\sqrt{64+36}=10\), so the radius is 5 m.
    3. \((x-4)^2+(y-3)^2=25\).
    4. \((4-4)^2+(8-3)^2=0+25=25\) ✓. Area \(=\pi\times5^2=\) 78.5 m².

15. Answer to Question 15

    1. \((5-1)^2+(4-2)^2=16+4=20\). The point satisfies the equation, so \(P\) is on the circle.
    2. Gradient \(=\dfrac{4-2}{5-1}=\dfrac12\). Use the centre \(C(1,2)\) and the point \(P(5,4)\).
    3. Tangent gradient \(=-2\). \(y-4=-2(x-5)\), so \(y=-2x+14\). The tangent is perpendicular to the radius, so its gradient is the negative reciprocal of \(\tfrac12\). Check: \(-2\times5+14=4\).
    4. \(y=0\): \(x=7\), so \(A(7,0)\). \(x=0\): \(y=14\), so \(B(0,14)\). Area \(=\tfrac12\times7\times14=\) 49 square units.

16. Answer to Question 16

    1. Circle: \(36+4=40\) and \(4+36=40\). Line: \(-6+8=2\) and \(-2+8=6\). ✓ A point lies on a graph when its coordinates satisfy the equation.
    2. \(AB=\sqrt{(6-2)^2+(2-6)^2}=\sqrt{32}=\) 5.66. Use Pythagoras with changes of 4 across and 4 down.
    3. \(M=\left(\dfrac{6+2}2,\dfrac{2+6}2\right)=(4,4)\). Gradient \(OM=1\) and the gradient of the line is \(-1\); \(1\times(-1)=-1\), so \(OM\perp AB\). The product of the gradients is \(-1\), so \(OM\) is perpendicular to the chord.
    4. \(OM=\sqrt{16+16}=\sqrt{32}=5.66\). Check: \(OA^2=OM^2+AM^2=32+8=40\). ✓ \(AM\) is half of \(AB\), so \(AM^2=8\). The radius squared, 40, equals \(32+8\).

17. Answer to Question 17

    1. Midpoint of \(AB=\left(\dfrac{-2+6}2,\dfrac{3+9}2\right)=\) \((2,6)\). The centre of a circle is the midpoint of any diameter.
    2. \(AB=\sqrt{(6+2)^2+(9-3)^2}=\sqrt{100}=10\), so the radius is 5. The radius is half of the diameter \(AB\), which we find with Pythagoras.
    3. \((x-2)^2+(y-6)^2=25\). Check with \(A\): \(16+9=25\). Use \((x-a)^2+(y-b)^2=r^2\) with centre \((2,6)\) and \(r^2=25\).
    4. \((2-2)^2+(11-6)^2=0+25=25\). ✓ The point satisfies the equation, so it lies on the circle.

18. Answer to Question 18

    1. \(PA^2=OP^2-OA^2=100-25=75\), so \(PA=\sqrt{75}=\) 8.66. \(OA\) is perpendicular to the tangent \(PA\), so triangle \(OAP\) is right-angled at \(A\) with hypotenuse \(OP=10\). Exactly, \(PA=5\sqrt3\).
    2. \(\sin(OPA)=\dfrac{5}{10}=0.5\), so \(OPA=\) \(30^\circ\). \(OA=5\) is opposite angle \(OPA\) and \(OP=10\) is the hypotenuse, so we use sine.
    3. By symmetry the tangents make equal angles with \(OP\), so \(APB=2\times30=60^\circ\). \(OP\) is a line of symmetry of the figure.
    4. Two right-angled triangles each of area \(\tfrac12\times5\times8.66=21.65\), so the area is 43.30 square units. The exact area is \(25\sqrt3\approx43.30\).

19. Answer to Question 19

    1. \(x^2+(2x+k)^2=5\) gives \(x^2+4x^2+4kx+k^2=5\), so \(5x^2+4kx+k^2-5=0\). Replace \(y\) by \(2x+k\) and expand \((2x+k)^2=4x^2+4kx+k^2\).
    2. \(b^2-4ac=(4k)^2-4\times5\times(k^2-5)=16k^2-20k^2+100=100-4k^2=0\), so \(k^2=25\). Here \(a=5\), \(b=4k\) and \(c=k^2-5\). A tangent meets the circle at one point, so the discriminant is zero.
    3. \(k=5\) or \(k=-5\). For \(k=5\): \(5x^2+20x+20=0\), so \(x^2+4x+4=0\), \(x=-2\) and \(y=2(-2)+5=1\): the point of contact is \((-2,1)\) (check: \(4+1=5\)). Take the square root of 25, remembering both signs. Substitute \(k=5\) and divide by 5: \((x+2)^2=0\), so there is one repeated root.

20. Answer to Question 20

    1. \(x^2-6x=(x-3)^2-9\) and \(y^2-8y=(y-4)^2-16\). So \((x-3)^2+(y-4)^2=25\): centre \((3,4)\), radius 5.
    2. At \((0,0)\): \(9+16=25\) ✓. Gradient of the radius \(=\dfrac43\), tangent gradient \(-\dfrac34\) through the origin: \(y=-\dfrac34x\).
    3. \(y=0\): \(x^2-6x=0\), so \(x=6\): \(A(6,0)\). \(x=0\): \(y^2-8y=0\), so \(y=8\): \(B(0,8)\). Angle \(AOB=90^\circ\) (the axes), and an angle of \(90^\circ\) at the circumference stands on a diameter. Check: \(AB=\sqrt{36+64}=10=2\times5\). ✓