Higher Maths: the straight line and the circle

Coordinate geometry at Higher runs on a small number of relationships used over and over: perpendicular gradients multiply to minus one, a median runs to the midpoint of the opposite side, a tangent is perpendicular to the radius at the point of contact, and the number of times a line meets a circle is decided by a discriminant.

Ten single-answer questions work through those. They cover perpendicular gradients, the gradient of a line from the angle it makes with the x-axis, the equation of a perpendicular line through a given point, the equation of a circle from its centre and radius, recovering the centre and radius from the general equation by completing the square, deciding whether a point lies inside a circle, the gradient of a tangent, the midpoint, the equation of a median, and testing a line for tangency.

Two questions reward checking rather than recognising. In the median question, two of the wrong lines pass through the vertex but have the wrong gradient and one has the right gradient but misses the vertex, so only a line satisfying both conditions survives. In the tangency question the working goes all the way to a discriminant of zero, which is what distinguishes a tangent from a chord.

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  • Find a perpendicular gradient as a negative reciprocal, doing both steps
  • Find the gradient of a line from the angle it makes with the positive x-axis
  • Write the equation of a line through a given point perpendicular to another
  • Write the equation of a circle from its centre and radius, remembering to square the radius
  • Recover the centre and radius from the general equation of a circle
  • Decide whether a point lies inside, on or outside a circle
  • Find the gradient of a tangent from the gradient of the radius at the point of contact
  • Find a midpoint and use it to write the equation of a median
  • Test a line against a circle using the discriminant of the resulting quadratic

How many points does the line $y = x + 4$ have in common with the circle $x^2 + y^2 = 8$? — Substituting gives $x^2 + 4x + 4 = 0$, whose discriminant is $16 - 16 = 0$: one repeated root, so the line is a tangent, touching at $(-2, 2)$.

Sample question

A straight line has a gradient of $\frac{2}{5}$. What is the gradient of any line perpendicular to it?

See the answer

$-\frac{5}{2}$

Perpendicular gradients multiply to $-1$, so the gradient wanted is the negative reciprocal of $\frac{2}{5}$, which is $-\frac{5}{2}$. Check: $\frac{2}{5} \times -\frac{5}{2} = -1$. The two closest wrong options each do half the job: $\frac{5}{2}$ inverts the fraction but keeps the sign positive, and $-\frac{2}{5}$ changes the sign but never inverts.

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