The equation of a circle with centre the origin and with a general centre, the general equation of a circle, finding the centre and radius, the intersection of a line and a circle, and the equation of a tangent to a circle.

What this dot point is asking

The SQA wants you to write the equation of a circle with any centre, read the centre and radius from the general equation, find where a line meets a circle, and find the equation of a tangent to a circle.

The equation of a circle

There are two standard forms. The centre-radius form is read directly off the equation, while the general (expanded) form hides the centre and radius inside its coefficients, so you must decode them.

A real circle needs $g^2 + f^2 - c > 0$, otherwise the radius is not a real number and the equation represents no points (or a single point if the value is zero).

Line and circle

To find where a line meets a circle, substitute the line equation into the circle equation. This produces a quadratic in one variable, and its roots are the x-coordinates of the intersection points. The discriminant of that quadratic classifies the geometry: two real roots mean the line cuts the circle, a repeated root means it is a tangent, and no real roots mean it misses.

If the discriminant of that quadratic is zero, the line is a tangent; if it is negative, the line misses the circle.

Tangents

A tangent is perpendicular to the radius at the point of contact, so its gradient is the negative reciprocal of the radius gradient. The method is to find the gradient of the radius from the centre to the point of contact, take the negative reciprocal for the tangent gradient, then use the point of contact in $y - y_1 = m(x - x_1)$.

Examples in context

Coordinate-circle work models the reach of a transmitter or the range of a sensor. If a mast at $(2, 3)$ has a signal range of $5$ km, every point it reaches satisfies $(x - 2)^2 + (y - 3)^2 \le 25$. A straight road $y = 2x - 1$ is within range exactly where substituting into $(x - 2)^2 + (y - 3)^2 = 25$ gives real solutions, so the discriminant of the resulting quadratic tells the planner whether and where the road enters coverage.

Try this

Q1. State the centre and radius of $x^2 + y^2 - 6x + 4y - 12 = 0$. [3 marks]

• Cue. $g = -3$, $f = 2$, $c = -12$; centre $(3, -2)$, radius $\sqrt{9 + 4 + 12} = 5$.

Q2. Show that the line $y = 2x + 5$ is a tangent to $x^2 + y^2 = 5$. [3 marks]

• Cue. Substituting gives $x^2 + (2x + 5)^2 = 5$, so $5x^2 + 20x + 20 = 0$; the discriminant $20^2 - 4(5)(20) = 0$, so the line touches once.

Q3. Find the equation of the circle with centre $(2, -1)$ that passes through $(5, 3)$. [3 marks]

• Cue. Radius $= \sqrt{(5 - 2)^2 + (3 + 1)^2} = \sqrt{9 + 16} = 5$, so the equation is $(x - 2)^2 + (y + 1)^2 = 25$.