Hard GCSE Equation of a circle Questions

Challenging, exam-style GCSE Equation of a circle questions with worked solutions. Stretch yourself on the hardest tangent to a circle, perpendicular gradients, equation of a line, negative coordinates problems.

tangent to a circleperpendicular gradientsequation of a linenegative coordinatesline-circle intersectionsubstitution
GCSE Higher34 questionsStep-by-step solutions
Question 1
6 markschallenging
The point (a,b)(a, b) lies on the circle x2+y2=r2x^2 + y^2 = r^2, where a≠0a \neq 0 and b≠0b \neq 0. Prove that the tangent to the circle at (a,b)(a, b) has equation ax+by=r2ax + by = r^2.
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Worked solution

  1. Set up

    x2+y2=r2,point (a, b) on the circlex^2 + y^2 = r^2, \quad \text{point } (a,\, b) \text{ on the circle}

    The centre is the origin O(0, 0) and the radius is r.

  2. Write down what is known

    a2+b2=r2a^2 + b^2 = r^2

    Because (a, b) lies on the circle, its coordinates satisfy the equation. This fact will be used at the very end.

  3. Find the gradient of the radius

    mradius=b−0a−0=bam_{\text{radius}} = \frac{b - 0}{a - 0} = \frac{b}{a}

    The radius joins O(0, 0) to (a, b). This needs a not equal to 0, which is given.

  4. State the circle property

    tangent⊥radius at the point of contact\text{tangent} \perp \text{radius at the point of contact}

    This is the key fact the whole proof rests on.

  5. Apply the perpendicular rule

    mtangent×ba=−1m_{\text{tangent}} \times \frac{b}{a} = -1

    The gradients of perpendicular lines have product -1. This needs b not equal to 0, which is also given.

  6. Find the tangent gradient

    mtangent=−abm_{\text{tangent}} = -\frac{a}{b}

    Rearranging gives the negative reciprocal of b/a.

  7. Use the point-gradient form

    y−b=−ab(x−a)y - b = -\frac{a}{b}(x - a)

    The tangent passes through (a, b) with gradient -a/b.

  8. Multiply through by b

    b(y−b)=−a(x−a)b(y - b) = -a(x - a)

    Clearing the fraction avoids messy algebra.

  9. Expand the left-hand side

    by−b2=−a(x−a)by - b^2 = -a(x - a)

    Multiply b into the bracket.

  10. Expand the right-hand side

    by−b2=−ax+a2by - b^2 = -ax + a^2

    -a times -a is +a2a^2.

  11. Collect the x and y terms

    ax+by=a2+b2ax + by = a^2 + b^2

    Add ax to both sides and add b2b^2 to both sides.

  12. Use the fact from step 2

    a2+b2=r2a^2 + b^2 = r^2

    (a, b) is on the circle, so the right-hand side is exactly r2r^2.

  13. Complete the proof

    ax+by=r2ax + by = r^2

    This is the required equation, so the result is proved.

  14. Test it

    (3, 4) on x2+y2=25⇒3x+4y=25(3,\, 4) \text{ on } x^2 + y^2 = 25 \Rightarrow 3x + 4y = 25

    Checking with a known example: the formula gives 3x+4y=253x + 4y = 25, which is the tangent found in the earlier questions.

  15. Note the special cases

    b=0⇒x=±r;a=0⇒y=±rb = 0 \Rightarrow x = \pm r; \qquad a = 0 \Rightarrow y = \pm r

    When b=0b = 0 the tangent is vertical and when a=0a = 0 it is horizontal — the gradient argument breaks down, which is exactly why the question excludes a=0a = 0 and b=0b = 0. Reassuringly, ax + by = r2r^2 still gives the right line in both cases.

Answer
Theradiusto(a,b)hasgradientb/a,sothetangenthasgradient−a/b.Usingy−b=−(a/b)(x−a)givesby−b2=−ax+a2,soax+by=a2+b2=r2,since(a,b)liesonthecircleThe radius to (a, b) has gradient b/a, so the tangent has gradient -a/b. Using y - b = -(a/b)(x - a) gives by - b^2 = -ax + a^2, so ax + by = a^2 + b^2 = r^2, since (a, b) lies on the circle
Question 2
6 markschallenging
The line y=−12x+cy = -\frac{1}{2}x + c is a tangent to the circle x2+y2=45x^2 + y^2 = 45, where c<0c < 0. Work out the value of cc.
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Worked solution

  1. State the tangent condition

    tangent⇒discriminant=0\text{tangent} \Rightarrow \text{discriminant} = 0

    A tangent meets the circle exactly once, so the resulting quadratic has a repeated root.

  2. Substitute the line into the circle

    x2+(−12x+c)2=45x^2 + \left(-\frac{1}{2}x + c\right)^2 = 45

    Replace y by -x/2 + c in x2+y2=45x^2 + y^2 = 45.

  3. Expand the bracket

    (−12x+c)2=14x2−cx+c2\left(-\frac{1}{2}x + c\right)^2 = \frac{1}{4}x^2 - cx + c^2

    The middle term is 2 x (-x/2) x c = -cx.

  4. Collect terms

    x2+14x2−cx+c2=45x^2 + \frac{1}{4}x^2 - cx + c^2 = 45

    Substitute the expansion back in.

  5. Simplify the x squared terms

    54x2−cx+(c2−45)=0\frac{5}{4}x^2 - cx + (c^2 - 45) = 0

    1+1/4=5/41 + 1/4 = 5/4.

  6. Clear the fraction

    5x2−4cx+(4c2−180)=05x^2 - 4cx + (4c^2 - 180) = 0

    Multiply every term by 4 to make the coefficients whole numbers.

  7. Identify the coefficients

    a=5,b=−4c,constant=4c2−180a = 5,\quad b = -4c,\quad \text{constant} = 4c^2 - 180

    Read them off carefully.

  8. Set the discriminant to zero

    (−4c)2−4(5)(4c2−180)=0(-4c)^2 - 4(5)(4c^2 - 180) = 0

    This is the condition for exactly one point of contact.

  9. Expand the square

    16c2−20(4c2−180)=016c^2 - 20(4c^2 - 180) = 0

    (−4c)2=16c2(-4c)^2 = 16c^2.

  10. Multiply out the bracket

    16c2−80c2+3600=016c^2 - 80c^2 + 3600 = 0

    -20 x (−180)=+3600(-180) = +3600.

  11. Simplify

    −64c2+3600=0-64c^2 + 3600 = 0

    Collect the c2c^2 terms.

  12. Rearrange

    64c2=3600⇒c2=360064=225464c^2 = 3600 \Rightarrow c^2 = \frac{3600}{64} = \frac{225}{4}

    Divide both sides by 64 and cancel.

  13. Take both roots

    c=±152c = \pm \frac{15}{2}

    sqrt(225/4) = 15/2, so c=7.5c = 7.5 or c=−7.5c = -7.5.

  14. Use the condition

    c<0⇒c=−152c < 0 \Rightarrow c = -\frac{15}{2}

    The question says c is negative, so reject the positive root.

  15. Check

    5x2+30x+45=0⇒(x+3)2=0⇒x=−3,  y=−65x^2 + 30x + 45 = 0 \Rightarrow (x + 3)^2 = 0 \Rightarrow x = -3,\; y = -6

    With c=−15/2c = -15/2 the quadratic reduces to x2+6x+9=0x^2 + 6x + 9 = 0, a perfect square. The contact point is (-3, -6) and (−3)2+(−6)2=9+36=45(-3)^2 + (-6)^2 = 9 + 36 = 45, so it is on the circle.

Answer
c=−152c = -\frac{15}{2}
Question 3
6 markschallenging
The tangent to the circle x2+y2=20x^2 + y^2 = 20 at the point P(4,2)P(4, 2) crosses the xx-axis at AA and the yy-axis at BB. Work out the exact length of ABAB.
Show worked solution

Worked solution

  1. Find the radius

    r2=20⇒r=20=25r^2 = 20 \Rightarrow r = \sqrt{20} = 2\sqrt{5}

    The circle x2+y2=20x^2 + y^2 = 20 has exact radius 2 sqrt(5), about 4.47.

  2. Check P lies on the circle

    42+22=16+4=20  ✓4^2 + 2^2 = 16 + 4 = 20 \;\checkmark

    P(4, 2) satisfies the circle equation.

  3. Find the gradient of the radius OP

    mOP=24=12m_{OP} = \frac{2}{4} = \frac{1}{2}

    Cancel 2/4 down to 1/2.

  4. Use the tangent-radius property

    tangent⊥radius\text{tangent} \perp \text{radius}

    The tangent at P meets OP at a right angle.

  5. Find the tangent gradient

    m=−2m = -2

    The negative reciprocal of 1/2 is -2 (shortcut: −x1/y1=−4/2=−2-x1/y1 = -4/2 = -2).

  6. Use the point-gradient form

    y−2=−2(x−4)y - 2 = -2(x - 4)

    The tangent passes through P(4, 2).

  7. Expand

    y−2=−2x+8y - 2 = -2x + 8

    -2 times -4 is +8.

  8. Rearrange

    y=−2x+10y = -2x + 10

    Add 2 to both sides.

  9. Check against the standard form

    4x+2y=20  ⇔  2x+y=10  ⇔  y=−2x+104x + 2y = 20 \;\Leftrightarrow\; 2x + y = 10 \;\Leftrightarrow\; y = -2x + 10

    x1 x + y1 y=r2y = r^2 gives 4x+2y=204x + 2y = 20, which is the same line.

  10. Find A on the x-axis

    y=0⇒0=−2x+10⇒x=5,A(5, 0)y = 0 \Rightarrow 0 = -2x + 10 \Rightarrow x = 5, \quad A(5,\, 0)

    Set y=0y = 0 in the tangent equation: the tangent crosses the x-axis at (5, 0).

  11. Find B on the y-axis

    x=0⇒y=10,B(0, 10)x = 0 \Rightarrow y = 10, \quad B(0,\, 10)

    Set x=0x = 0 in the tangent equation: the tangent crosses the y-axis at (0, 10).

  12. Spot the right angle at O

    OA lies on the x-axis,OB lies on the y-axisOA \text{ lies on the x-axis},\quad OB \text{ lies on the y-axis}

    The axes meet at 90 degrees, so triangle OAB is right-angled at the origin and Pythagoras applies.

  13. Use Pythagoras on triangle OAB

    AB2=52+102=25+100=125AB^2 = 5^2 + 10^2 = 25 + 100 = 125

    AB is the hypotenuse, with legs OA = 5 and OB = 10.

  14. Simplify the surd

    AB=125=25×5=55AB = \sqrt{125} = \sqrt{25 \times 5} = 5\sqrt{5}

    25 is the largest square factor of 125.

  15. State the exact length

    AB=55AB = 5\sqrt{5}

    The exact length is 555\sqrt{5}, about 11.18. (Note OA×OB=50=2×r2×…OA \times OB = 50 = 2 \times r^2 \times \ldots — but the surd form is what is asked for.)

Answer
555\sqrt{5}
Question 4
6 markschallenging
The tangent to the circle x2+y2=25x^2 + y^2 = 25 at the point A(3,4)A(3, 4) and the tangent at the point B(5,0)B(5, 0) meet at the point CC. Work out the coordinates of CC.
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Worked solution

  1. Find the radius

    r2=25⇒r=5r^2 = 25 \Rightarrow r = 5

    The circle x2+y2=25x^2 + y^2 = 25 has centre O(0, 0) and radius 5.

  2. Check A lies on the circle

    32+42=25  ✓3^2 + 4^2 = 25 \;\checkmark

    A(3, 4) is on the circle.

  3. Check B lies on the circle

    52+02=25  ✓5^2 + 0^2 = 25 \;\checkmark

    B(5, 0) is on the circle — it is where the circle meets the positive x-axis.

  4. Find the gradient of OA

    mOA=43m_{OA} = \frac{4}{3}

    The radius from the origin to A.

  5. Find the tangent gradient at A

    mA=−34m_A = -\frac{3}{4}

    The negative reciprocal of 4/3.

  6. Write the tangent at A

    y−4=−34(x−3)y - 4 = -\frac{3}{4}(x - 3)

    Point-gradient form through A(3, 4).

  7. Tidy the tangent at A

    4y−16=−3x+9  ⇒  3x+4y=254y - 16 = -3x + 9 \;\Rightarrow\; 3x + 4y = 25

    Multiply by 4 and collect: this is x1 x + y1 y=r2y = r^2.

  8. Look at the radius to B

    the radius OB runs along the x-axis\text{the radius } OB \text{ runs along the x-axis}

    B(5, 0) is level with the centre, so OB is horizontal with gradient 0.

  9. Deal with the special case

    y1=0⇒−x1y1 is undefinedy_1 = 0 \Rightarrow -\frac{x_1}{y_1} \text{ is undefined}

    The formula -x1/y1 breaks down because we cannot divide by 0 — this is a warning that the tangent is VERTICAL, not that something has gone wrong.

  10. Write the tangent at B

    x=5x = 5

    A line perpendicular to a horizontal radius is vertical, so the tangent at B is the vertical line x=5x = 5. It has no gradient and cannot be written as y = mx + c.

  11. Substitute into the tangent at A

    3(5)+4y=253(5) + 4y = 25

    C lies on both tangents, so put x=5x = 5 into 3x+4y=253x + 4y = 25.

  12. Simplify

    15+4y=2515 + 4y = 25

    3 times 5 is 15.

  13. Solve for y

    4y=10⇒y=104=524y = 10 \Rightarrow y = \frac{10}{4} = \frac{5}{2}

    Cancel 10/4 down to 5/2.

  14. State C

    C(5, 52)C\left(5,\, \frac{5}{2}\right)

    The two tangents cross at (5, 2.5).

  15. Check

    CA=CB:  CB=52 and CA=(5−3)2+(52−4)2=4+94=52CA = CB: \; CB = \frac{5}{2} \text{ and } CA = \sqrt{(5-3)^2 + \left(\frac{5}{2}-4\right)^2} = \sqrt{4 + \frac{9}{4}} = \frac{5}{2}

    The two tangent lengths from an external point are always equal, and both come to 5/2 — a strong check that C is right.

Answer
(5, 52)\left(5,\, \frac{5}{2}\right)
Question 5
5 markschallenging
The points A(1,8)A(1, 8) and B(7,4)B(7, 4) lie on the circle x2+y2=65x^2 + y^2 = 65. Work out the exact length of the chord ABAB.
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Worked solution

  1. Find the radius

    r2=65⇒r=65r^2 = 65 \Rightarrow r = \sqrt{65}

    65=5×1365 = 5 \times 13 has no square factors, so the radius stays as 65\sqrt{65}, about 8.06.

  2. Check A lies on the circle

    12+82=1+64=65  ✓1^2 + 8^2 = 1 + 64 = 65 \;\checkmark

    A(1, 8) satisfies x2+y2=65x^2 + y^2 = 65.

  3. Check B lies on the circle

    72+42=49+16=65  ✓7^2 + 4^2 = 49 + 16 = 65 \;\checkmark

    B(7, 4) satisfies it too — 65 can be written as a sum of two squares in more than one way.

  4. Recall the distance formula

    AB=(x2−x1)2+(y2−y1)2AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

    This is Pythagoras applied to the horizontal and vertical gaps.

  5. Find the horizontal gap

    Δx=7−1=6\Delta x = 7 - 1 = 6

    The x-coordinate increases by 6 from A to B.

  6. Find the vertical gap

    Δy=4−8=−4\Delta y = 4 - 8 = -4

    The y-coordinate decreases by 4.

  7. Square the horizontal gap

    62=366^2 = 36

    Squaring the change in x.

  8. Square the vertical gap

    (−4)2=16(-4)^2 = 16

    Squaring removes the minus sign.

  9. Add them

    AB2=36+16=52AB^2 = 36 + 16 = 52

    The square of the chord length is 52.

  10. Take the square root

    AB=52AB = \sqrt{52}

    The exact length is the square root of 52.

  11. Find the largest square factor

    52=4×1352 = 4 \times 13

    4 is a perfect square and 13 is prime.

  12. Split the surd

    52=4×13\sqrt{52} = \sqrt{4} \times \sqrt{13}

    Separate the square factor.

  13. Simplify

    52=213\sqrt{52} = 2\sqrt{13}

    sqrt(4) = 2 comes out in front.

  14. Sense-check against the diameter

    213≈7.21and265≈16.122\sqrt{13} \approx 7.21 \quad\text{and}\quad 2\sqrt{65} \approx 16.12

    A chord must be shorter than the diameter, and 7.21<16.127.21 < 16.12, so the answer is sensible.

  15. State the exact length

    AB=213AB = 2\sqrt{13}

    The exact chord length is 2 sqrt(13).

Answer
2132\sqrt{13}

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