Hard GCSE Simultaneous (linear/quadratic) Questions

Challenging, exam-style GCSE Simultaneous (linear/quadratic) questions with worked solutions. Stretch yourself on the hardest substitution, general quadratic curve, paired solutions, circle problems.

substitutiongeneral quadratic curvepaired solutionscirclerearranging the line firstnon-monic quadratic
GCSE Higher34 questionsStep-by-step solutions
Question 1
6 markschallenging
The line y=2x+ky = 2x + k crosses the circle x2+y2=5x^2 + y^2 = 5 at two distinct points. Find the range of values of kk.
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Worked solution

  1. Substitute the line into the circle equation

    x2+(2x+k)2=5x^2 + \left(2x + k\right)^2 = 5

    Substitution leaves a quadratic in xx whose coefficients involve kk.

  2. Expand the bracket

    x2+4x2+4kx+k2=5x^2 + 4x^{2} + 4kx + k^{2} = 5

    Use (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2 with b=kb = k.

  3. Rearrange so the quadratic equals zero

    5x2+4kx+(k2−5)=05x^{2} + 4 kx + \left(k^{2} - 5\right) = 0

    Collect like terms and take the constant across.

  4. Write down a, b and c

    a=5,b=4k,c=k2−5a = 5,\quad b = 4 k,\quad c = k^{2} - 5

    Both bb and cc contain the unknown kk.

  5. "Two distinct points" means two distinct real roots

    b2−4ac>0b^2 - 4ac > 0

    Each real root of the quadratic gives one point of intersection, so two distinct points require a strictly positive discriminant.

  6. Write down the discriminant

    b2−4ac=(4k)2−4(5)(k2−5)b^2 - 4ac = \left(4 k\right)^2 - 4\left(5\right)\left(k^{2} - 5\right)

    Substitute the coefficients.

  7. Simplify

    =100−4k2= 100 - 4 k^{2}

    The k2k^2 terms partly cancel, leaving a quadratic in kk.

  8. Form the inequality

    100−4k2>0100 - 4 k^{2} > 0

    This is the condition for two distinct intersection points.

  9. Rearrange

    k2<25k^2 < 25

    Divide by the negative coefficient of k2k^2 and REVERSE the inequality sign — the classic slip here.

  10. Take the square root

    −5<k<5-5 < k < 5

    k2<25k^2 < 25 means ∣k∣<5|k| < 5, which is the double inequality −5<k<5-5 < k < 5.

  11. Check the boundary values

    k=±5: b2−4ac=0k = \pm 5: \ b^2 - 4ac = 0

    At the boundaries the discriminant is zero: the line is a tangent, touching the circle at one point rather than cutting it at two. So the boundaries are excluded.

  12. Check a value inside the range

    k=0: b2−4ac=100>0 ✓k = 0: \ b^2-4ac = 100 > 0 \ \checkmark

    The line y=2xy = 2x passes through the centre of the circle, so of course it cuts it twice.

  13. Check a value outside the range

    k=6: b2−4ac=−44<0 ✓k = 6: \ b^2-4ac = -44 < 0 \ \checkmark

    A negative discriminant means the line misses the circle entirely — outside the required range, as expected.

  14. Interpret geometrically

    tangents at k=−5 and k=5\text{tangents at } k = -5 \text{ and } k = 5

    As kk increases the line slides upwards. Between the two tangent positions it cuts the circle twice; outside them it misses.

  15. State the range

    −5<k<5-5 < k < 5

    The line cuts the circle at two distinct points exactly when −5<k<5-5 < k < 5.

Answer
−5<k<5-5 < k < 5
Question 2
6 markschallenging
The line y=3x−5y = 3x - 5 cuts the circle x2+y2=25x^2 + y^2 = 25 at the points AA and BB. Work out the exact length of ABAB.
Show worked solution

Worked solution

  1. Label the two equations

    (1) x2+y2=25(2) y=3x−5(1)\ x^2 + y^2 = 25 \qquad (2)\ y = 3 x - 5

    One equation is quadratic and one is linear, so you cannot eliminate a variable by adding or subtracting. Substitution is the only route.

  2. Substitute the line into the circle equation

    x2+(3x−5)2=25x^2 + \left(3 x - 5\right)^2 = 25

    Replace yy in the circle equation by the expression from the line. Every yy disappears, leaving one equation in xx alone.

  3. Expand the bracket

    x2+9x2−30x+25=25x^2 + 9 x^{2} - 30 x + 25 = 25

    Use (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2 — squaring each term separately is the classic slip here.

  4. Rearrange so the quadratic equals zero

    10x2−30x=010 x^{2} - 30 x = 0

    Collect every term on one side. A quadratic can only be factorised once it is equal to zero.

  5. Divide every term by 10

    x2−3x=0x^{2} - 3 x = 0

    All three coefficients are multiples of 1010, so dividing through keeps the same solutions but makes the factorising much easier.

  6. Factorise

    x(x−3)=0x(x - 3) = 0

    Look for the pair of numbers that multiply to give the constant term and add to give the coefficient of xx.

  7. Solve for x

    x=0orx=3x = 0 \quad \text{or} \quad x = 3

    These are the xx-coordinates of the two points where the line meets the curve.

  8. Find the y that goes with the first x-value

    y=3(0)−5=−5y = 3 \left(0\right) - 5 = -5

    Substitute this xx into the LINEAR equation — it is far less work than the quadratic, and it gives the yy that belongs to this xx.

  9. Find the y that goes with the second x-value

    y=3(3)−5=4y = 3 \left(3\right) - 5 = 4

    Substitute this xx into the LINEAR equation — it is far less work than the quadratic, and it gives the yy that belongs to this xx.

  10. Name the two points

    A(0, −5),B(3, 4)A\left(0,\ -5\right), \quad B\left(3,\ 4\right)

    Each xx carries the yy found with it. AA and BB are the two ends of the chord.

  11. Check both points lie on the circle

    (0)2+(−5)2=25 ✓(3)2+(4)2=25 ✓\left(0\right)^2 + \left(-5\right)^2 = 25 \ \checkmark \qquad \left(3\right)^2 + \left(4\right)^2 = 25 \ \checkmark

    Both pairs satisfy the circle equation as well as the line, so the pairing is right.

  12. Write down the distance formula

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

    The length of the chord is just the distance between its two endpoints — Pythagoras on the coordinate differences.

  13. Work out the differences

    x2−x1=3,y2−y1=9x_2 - x_1 = 3, \qquad y_2 - y_1 = 9

    Subtract in the same order for both coordinates. The signs do not matter once they are squared.

  14. Substitute into the distance formula

    AB=(3)2+(9)2=90AB = \sqrt{\left(3\right)^2 + \left(9\right)^2} = \sqrt{90}

    Square each difference and add.

  15. Simplify the surd

    AB=90=310AB = \sqrt{90} = 3\sqrt{10}

    90\sqrt{90} simplifies by taking out the largest square factor. Leave the answer exact — do not round.

Answer
AB=310AB = 3\sqrt{10}
Question 3
5 markschallenging
Which one of these lines meets the curve y=x2y = x^2 at exactly one point?
Show worked solution

Worked solution

  1. Set up the general test

    y=x2 and y=4x+c ⇒ x2−4x−c=0y = x^2 \ \text{and} \ y = 4x + c \ \Rightarrow \ x^2 - 4x - c = 0

    Every option has gradient 44, so substitute the general line y=4x+cy = 4x + c into the curve. Only the constant changes.

  2. Write down the discriminant in terms of c

    b2−4ac=(−4)2−4(1)(−c)=16+4cb^2 - 4ac = (-4)^2 - 4(1)(-c) = 16 + 4c

    The number of intersection points depends only on the sign of 16+4c16 + 4c.

  3. State the condition for exactly one point

    16+4c=0 ⇒ c=−416 + 4c = 0 \ \Rightarrow \ c = -4

    Exactly one point of intersection means a repeated root, which means a discriminant of zero.

  4. Test the line y=4x−4y = 4x - 4

    x2−4x+4=0,b2−4ac=0 ⇒ 1 pointx^{2} - 4 x + 4 = 0, \quad b^2-4ac = 0 \ \Rightarrow \ 1 \text{ point}

    This is the tangent: a zero discriminant, one repeated root, one point of contact.

  5. Test the line y=4x−3y = 4x - 3

    x2−4x+3=0,b2−4ac=4 ⇒ 2 pointsx^{2} - 4 x + 3 = 0, \quad b^2-4ac = 4 \ \Rightarrow \ 2 \text{ point}s

    A positive discriminant, so this line cuts the curve twice — not the answer.

  6. Test the line y=4x−5y = 4x - 5

    x2−4x+5=0,b2−4ac=−4 ⇒ 0 pointsx^{2} - 4 x + 5 = 0, \quad b^2-4ac = -4 \ \Rightarrow \ 0 \text{ point}s

    A negative discriminant, so this line misses the curve completely.

  7. Test the line y=4xy = 4x

    x2−4x=0,b2−4ac=16 ⇒ 2 pointsx^{2} - 4 x = 0, \quad b^2-4ac = 16 \ \Rightarrow \ 2 \text{ point}s

    A positive discriminant, so this line cuts the curve twice — not the answer.

  8. Test the line y=4x+4y = 4x + 4

    x2−4x−4=0,b2−4ac=32 ⇒ 2 pointsx^{2} - 4 x - 4 = 0, \quad b^2-4ac = 32 \ \Rightarrow \ 2 \text{ point}s

    A positive discriminant, so this line cuts the curve twice — not the answer.

  9. Collect the results

    c=−4→1,c=−3→2,c=−5→0,c=0→2,c=4→2c = -4 \rightarrow 1, \quad c = -3 \rightarrow 2, \quad c = -5 \rightarrow 0, \quad c = 0 \rightarrow 2, \quad c = 4 \rightarrow 2

    Only one of the five lines gives a single point of intersection: the one with c=−4c = -4.

  10. Identify the answer

    y=4x−4y = 4x - 4

    This is the only line of the five whose discriminant is zero, so it is the only tangent.

  11. Find the point of contact for the tangent

    x2−4x+4=0 ⇒ (x−2)2=0 ⇒ x=2x^2 - 4x + 4 = 0 \ \Rightarrow \ (x-2)^2 = 0 \ \Rightarrow \ x = 2

    The repeated root is x=2x = 2.

  12. Find the y-coordinate

    y=4(2)−4=4y = 4(2) - 4 = 4

    Substituting into the line gives y=4y = 4; the curve agrees, since 22=42^2 = 4.

  13. Check the point on both graphs

    22=4 ✓4(2)−4=4 ✓2^2 = 4 \ \checkmark \qquad 4(2) - 4 = 4 \ \checkmark

    The point (2,4)(2, 4) lies on the curve and on the line, and it is the only such point.

  14. Why the others fail

    c>−4⇒2 points,c<−4⇒0 pointsc > -4 \Rightarrow 2 \text{ points}, \qquad c < -4 \Rightarrow 0 \text{ points}

    Any line y=4x+cy = 4x + c with c>−4c > -4 cuts the curve twice; with c<−4c < -4 it misses it. Only c=−4c = -4 touches.

  15. State the answer

    y=4x−4 touches y=x2 at (2, 4)y = 4x - 4 \ \text{touches } y = x^2 \text{ at } (2,\ 4)

    It is the only one of the five lines that meets the curve exactly once.

Answer
(2, 4) on y=4x−4(2,\ 4) \ \text{on} \ y = 4x - 4
Question 4
5 markschallenging
A student solves y=x2y = x^2 and y=x+6y = x + 6 and correctly gets x=−2x = -2 and x=3x = 3. The student then writes: "The solutions are (−2,9)(-2, 9) and (3,4)(3, 4)." Explain the mistake and give the correct solutions.
Show worked solution

Worked solution

  1. Label the two equations

    (1) y=x2(2) y=x+6(1)\ y = x^{2} \qquad (2)\ y = x + 6

    One equation is quadratic and one is linear, so you cannot eliminate a variable by adding or subtracting. Substitution is the only route.

  2. Substitute the line into the curve

    x2=x+6x^{2} = x + 6

    Both equations give yy, so where the graphs meet the two expressions for yy are equal. That leaves one equation in xx alone.

  3. Rearrange so the quadratic equals zero

    x2−x−6=0x^{2} - x - 6 = 0

    Collect every term on one side. A quadratic can only be factorised once it is equal to zero.

  4. Factorise

    (x+2)(x−3)=0(x + 2)(x - 3) = 0

    Look for the pair of numbers that multiply to give the constant term and add to give the coefficient of xx.

  5. Solve for x

    x=−2orx=3x = -2 \quad \text{or} \quad x = 3

    These are the xx-coordinates of the two points where the line meets the curve.

  6. Test the student’s first point in the curve

    (−2, 9): y=(−2)2=4≠9\left(-2,\ 9\right): \ y = \left(-2\right)^2 = 4 \ne 9

    The student paired x=−2x = -2 with y=9y = 9. But −22=4-2^2 = 4, not 99, so this point is not on the curve y=x2y = x^2 at all.

  7. Test the student’s first point in the line

    (−2, 9): y=(−2)+6=4≠9\left(-2,\ 9\right): \ y = \left(-2\right) + 6 = 4 \ne 9

    It fails the line as well. A point that satisfies neither equation cannot be a solution.

  8. Diagnose the mistake

    x=−2→y=4,x=3→y=9x = -2 \rightarrow y = 4, \qquad x = 3 \rightarrow y = 9

    The two xx-values are right, but the yy-values have been swapped. Solving the quadratic only gives you half of each solution.

  9. Find the correct y for the first root

    x=(−2): y=(−2)+6=4x = \left(-2\right): \ y = \left(-2\right) + 6 = 4

    Substitute each xx into the LINEAR equation and keep the answer with that xx.

  10. Find the correct y for the second root

    x=(3): y=(3)+6=9x = \left(3\right): \ y = \left(3\right) + 6 = 9

    The second xx carries the second yy — they belong together.

  11. Check the first correct point in both equations

    y=(−2)2=4 ✓y=(−2)+6=4 ✓y = \left(-2\right)^2 = 4 \ \checkmark \qquad y = \left(-2\right) + 6 = 4 \ \checkmark

    (-2, 4) satisfies BOTH equations, so it is a genuine solution.

  12. Check the second correct point in both equations

    y=(3)2=9 ✓y=(3)+6=9 ✓y = \left(3\right)^2 = 9 \ \checkmark \qquad y = \left(3\right) + 6 = 9 \ \checkmark

    (3, 9) also satisfies both equations.

  13. State the rule

    each x keeps the y found with it\text{each } x \text{ keeps the } y \text{ found with it}

    A solution of a pair of simultaneous equations is a PAIR. Always substitute each xx back to get its own yy, then check the pair in both equations.

  14. Interpret the answer geometrically

    2 real solution pairs⇒2 points of intersection\text{2 real solution pairs} \Rightarrow \text{2 points of intersection}

    There are two solution pairs, so the line cuts the curve at two points. Solving the pair algebraically and reading the graph give the same information.

  15. State the correct solutions

    (−2, 4)and(3, 9)\left(-2,\ 4\right) \quad \text{and} \quad \left(3,\ 9\right)

    The correct solutions are (-2, 4) and (3, 9) — the student had the right xx-values but the wrong pairing.

Answer
(−2, 4)and(3, 9)\left(-2,\ 4\right) \quad \text{and} \quad \left(3,\ 9\right)
Question 5
6 markschallenging
The line y=2x+cy = 2x + c is a tangent to the circle x2+y2=5x^2 + y^2 = 5, where cc is positive. Work out the coordinates of the point where the line touches the circle.
Show worked solution

Worked solution

  1. Substitute the line into the circle equation

    x2+(2x+c)2=5x^2 + \left(2x + c\right)^2 = 5

    Replace yy by 2x+c2x + c. Both xx and the unknown cc remain.

  2. Expand the bracket

    x2+4x2+4cx+c2=5x^2 + 4x^{2} + 4cx + c^{2} = 5

    Use (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2 with b=cb = c.

  3. Rearrange so the quadratic equals zero

    5x2+4cx+(c2−5)=05x^{2} + 4 cx + \left(c^{2} - 5\right) = 0

    Collect like terms. Careful: the "cc" of the quadratic formula is c2−5c^{2} - 5, not the cc in the equation of the line.

  4. A tangent means one repeated root

    b2−4ac=0b^2 - 4ac = 0

    The line touches the circle exactly once, so the quadratic has a repeated root.

  5. Write down the discriminant

    (4c)2−4(5)(c2−5)=100−4c2\left(4 c\right)^2 - 4\left(5\right)\left(c^{2} - 5\right) = 100 - 4 c^{2}

    Substitute the coefficients and simplify.

  6. Set the discriminant equal to zero

    100−4c2=0100 - 4 c^{2} = 0

    An equation in cc alone.

  7. Solve for c

    c2=25 ⇒ c=±5c^2 = 25 \ \Rightarrow \ c = \pm 5

    There are two tangents with gradient 22, one either side of the circle.

  8. Choose the positive value

    c=5c = 5

    The question says cc is positive, so take the positive root.

  9. Substitute c back into the quadratic

    5x2+20x+20=05 x^{2} + 20 x + 20 = 0

    Now every coefficient is a number, so the repeated root can be found.

  10. Divide through by the common factor

    x2+4x+4=0x^{2} + 4 x + 4 = 0

    Dividing by the leading coefficient makes the perfect square obvious.

  11. Factorise the perfect square

    (x+2)2=0\left(x + 2\right)^2 = 0

    A repeated factor — exactly what a tangent produces.

  12. Solve for x

    x=−2x = -2

    One value of xx only: the line touches the circle at a single point.

  13. Find y from the line

    y=2(−2)+5=1y = 2 \left(-2\right) + 5 = 1

    Substitute the repeated root into the line to get the yy that belongs to it.

  14. Check the point lies on the circle

    (−2)2+(1)2=5 ✓\left(-2\right)^2 + \left(1\right)^2 = 5 \ \checkmark

    The point satisfies the circle equation as well as the line, so it really is the point of contact.

  15. State the point of contact

    (−2, 1)\left(-2,\ 1\right)

    With c=5c = 5 the line touches the circle at (-2, 1).

Answer
(−2, 1)\left(-2,\ 1\right)

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