Bradley's Maths

Straight Line Intersecting a Circle

GCSE all examination boards and IGCSE Cambridge (0580)

Exam Question

The straight line $y=2x-3$ intersects the circle $x^2+y^2=41$ in two points, $A$ and $B$.

Find the length of the chord, $AB$.

In this question we need to find the length of the chord connecting two points $A$ and $B$. But to do that, and without being told, we need to find the coordinates of the points of intersection between the given straight line and the given circle. With very little scaffolding to guide the student through the process this is a GCSE Grade 9 or IGCSE A* style question which only the best students will get full marks on.

The truth is... it isn't as difficult as it first appears!

I challenge you now, before you look through the solution or watch the video, to grab a sheet of paper and a pencil and try it for yourself.

Solution

Substitute the value of $y$ into the equation of the circle:

$$x^2+(2x-3)^2=41$$

Expand the left hand side (LHS):

$$x^2+4x^2-12x+9=41$$

Subtract 41 from each side and simplify the LHS:

$$5x^2-12x-32=0$$

Factorise the LHS:

$$5\times(-32)=-160$$ $$\text{factors of }160\text{ that sum to }-12\text{ are }-20\text{ and }8$$ \begin{align*} 5x^2-12x-32 &= 0 \\ \equiv 5x^2-20x+8x-32 &= 0 \\ 5x(x-4)+8(x-4) &= 0 \\ (5x+8)(x-4) &= 0 \end{align*}

We now have our solutions for $x$:

$$x=-\frac{8}{5}\text{ and }x=4$$

And we substitute these values into our original linear equation to find the corresponding $y$ values:

$$y=2\left(-\frac{8}{5}\right)-3\text{ and }y=2(4)-3$$

So our points of intersection are:

$$\left(-\frac{8}{5},-\frac{31}{5}\right) \text{ and } (4,5)$$

The difficult algebra is now finished. All that is left is a simple application of Pythagoras theorem to find the magnitude of $AB$

$$\lvert AB\rvert=\sqrt{\left(4+\frac{8}{5}\right)^2+\left(5+\frac{31}{5}\right)^2}$$ $$=12.5\text{ (3sf)}$$
The Head Teacher's Eye: Answer the whole question!

In questions such as this, involving often difficult algebra, students can easily lose marks by losing sight of the question. They work out the quadratic equation and come up with the solutions then either forget to apply the solutions to find the $y$-values or they mix-up the coordinate pairs putting the worong $x$ and $y$-values together.

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Topics covered in this question

  • Equation of a circle
  • Simultaneous equations
  • Factorising quadratics
  • Distance between two points

Video Explanation