Circles
One centre, one radius — the simplest curve in the plane, and a complete toolkit for tangents, chords and the way circles meet lines and one another
- The centre–radius form \((x-h)^2+(y-k)^2=r^2\) and how to read it both ways.
- The general equation \(x^2+y^2+2gx+2fy+c=0\), with centre \((-g,-f)\) and radius \(\sqrt{g^2+f^2-c}\).
- The diameter and parametric forms, and the circle through three points.
- Where a point sits relative to a circle, and the length of the tangent from it.
- How a line meets a circle, the tangency condition, and the equation of a tangent and the chord of contact.
- The chord with a given midpoint, families of circles, the radical axis, and orthogonal circles.
The Equation of a Circle
A circle is the set of all points at a fixed distance \(r\) (the radius) from a fixed point \((h,k)\) (the centre). Writing that distance condition with the distance formula gives the equation directly.
With centre at the origin this is simply \(x^2+y^2=r^2\). The form is reversible: given the equation you read off the centre and radius, and given a centre and radius you write the equation.
The General Equation
Expanding the centre–radius form and relabelling the constants gives the general equation of a circle. Any second-degree equation with equal \(x^2\) and \(y^2\) coefficients and no \(xy\) term is a circle.
Complete the square to recover the centre–radius form. The radius is real only when \(g^2+f^2-c>0\); it is a single point circle when \(g^2+f^2-c=0\), and imaginary (no real points) when \(g^2+f^2-c<0\).
If the equation comes as \(ax^2+ay^2+2gx+2fy+c=0\) with \(a\neq1\), divide through by \(a\) first. The centre and radius formulas only apply when the coefficients of \(x^2\) and \(y^2\) are both exactly \(1\).
Diameter & Parametric Forms
Two more descriptions are often the quickest route in. If you know the ends of a diameter, the diameter form writes the circle in one line; and the parametric form lets a single angle sweep out every point.
The diameter form uses the fact that any point on the circle sees a diameter at a right angle (Thales). The parametric form traces the circle as \(\theta\) runs from \(0\) to \(2\pi\), and is ideal for points written as \((h+r\cos\theta,\,k+r\sin\theta)\).
A Point and a Circle
For a circle \(S\equiv x^2+y^2+2gx+2fy+c=0\), substituting a point's coordinates gives a single number that tells you where the point lies. Write \(S_1\) for the value at \((x_1,y_1)\).
\(S_1<0\): the point lies strictly inside the circle.
\(S_1=0\): the point lies on the circle itself.
\(S_1>0\): the point lies outside, and tangents can be drawn.
From an external point, the two tangents to the circle have equal length \(\sqrt{S_1}\). This is just Pythagoras on the right triangle formed by the point, the centre, and the point of contact.
A Line and a Circle
Compare the perpendicular distance \(d\) from the centre to a line with the radius \(r\). That single comparison decides whether the line cuts, touches, or misses the circle.
For the line \(y=mx+k\) and circle \(x^2+y^2=a^2\), tangency \(d=r\) becomes the clean algebraic condition \(k^2=a^2(1+m^2)\).
Tangents & the Chord of Contact
The tangent at a point of the circle is perpendicular to the radius there. Replacing \(x^2\to xx_1\), \(y^2\to yy_1\), \(x\to\tfrac{x+x_1}{2}\) and \(y\to\tfrac{y+y_1}{2}\) in \(S\) produces the quantity \(T\), which packages all of these results.
The first, \(T=0\), is the tangent at a point \((x_1,y_1)\) on the circle. The second gives the two tangents of slope \(m\) to \(x^2+y^2=a^2\). From an external point, the same equation \(T=0\) is the chord of contact joining the two points of tangency.
Chords with a Given Midpoint
A chord is fixed once you know its midpoint, because the line from the centre to the midpoint is perpendicular to the chord. The equation drops straight out of \(T\) and \(S_1\).
The chord of \(S=0\) whose midpoint is \((x_1,y_1)\) is \(T=S_1\), where \(T\) is evaluated as above and \(S_1\) is the value of \(S\) at the midpoint. The normal at any point passes through the centre, so it is just the line through that point and \((-g,-f)\).
Families, Radical Axis & Orthogonality
Whole collections of circles can be handled with a single parameter, and two circles relate to each other through one special line and one neat angle condition.
Subtracting two circle equations cancels the \(x^2+y^2\) terms, leaving the straight radical axis — every point on it has equal tangent length to both circles. Two circles cut orthogonally (at right angles) exactly when \(2g_1g_2+2f_1f_2=c_1+c_2\).
Putting It to Work
Problem. Find the centre and radius of \(x^2+y^2-4x+6y-12=0\).
Solution. Here \(2g=-4,\ 2f=6,\ c=-12\), so \(g=-2,\ f=3\):
Problem. Find the circle with \((1,2)\) and \((5,8)\) as the ends of a diameter.
Solution. Use the diameter form \((x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0\):
Problem. Find the length of the tangent from \((7,9)\) to \(x^2+y^2-4x+6y-12=0\).
Solution. The length is \(\sqrt{S_1}\); substitute the point:
Problem. Find the tangent to \(x^2+y^2=25\) at the point \((3,4)\).
Solution. For \(x^2+y^2=a^2\) the tangent at \((x_1,y_1)\) is \(xx_1+yy_1=a^2\):
Problem. Find the tangents to \(x^2+y^2=4\) with slope \(1\).
Solution. Use \(y=mx\pm a\sqrt{1+m^2}\) with \(a=2,\ m=1\):
Problem. Find the chord of \(x^2+y^2=25\) whose midpoint is \((2,1)\).
Solution. Use \(T=S_1\), i.e. \(xx_1+yy_1-25=x_1^2+y_1^2-25\):
Chapter Summary
\((x-h)^2+(y-k)^2=r^2\); origin-centred \(x^2+y^2=r^2\).
Centre \((-g,-f)\), radius \(\sqrt{g^2+f^2-c}\) (real if \(>0\)).
Diameter \((x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0\); parametric \((h+r\cos\theta,\,k+r\sin\theta)\).
\(S_1\) signs give inside/on/outside; tangent length \(\sqrt{S_1}\); tangency \(d=r\).
Tangent / chord of contact \(T=0\); chord with midpoint \(T=S_1\).
Radical axis \(S_1-S_2=0\); orthogonal if \(2g_1g_2+2f_1f_2=c_1+c_2\).
Problems
Put each circle in general form first, then reach for \(S_1\), \(T\), or a distance comparison as the question demands. Difficulty rises down the list.
- Find the centre and radius of \(x^2+y^2-6x+8y-11=0\).
- Find the equation of the circle with centre \((-1,2)\) and radius \(3\).
- Find the circle with \((3,4)\) and \((-1,2)\) as the ends of a diameter.
- Determine whether \((1,2)\) lies inside, on, or outside \(x^2+y^2-4x-6y+9=0\).
- Find the length of the tangent from \((7,7)\) to \(x^2+y^2-4x-6y-12=0\).
- Find the tangent to \(x^2+y^2=13\) at the point \((2,3)\).
- Find the tangents to \(x^2+y^2=9\) having slope \(2\).
- For what value of \(c\) is the line \(y=2x+c\) a tangent to \(x^2+y^2=5\)?
- Find the chord of \(x^2+y^2=20\) whose midpoint is \((2,-1)\).
- Find the radical axis of \(x^2+y^2-4x-6y+3=0\) and \(x^2+y^2+2x-4y-5=0\).
- Find the equation of the circle passing through \((0,0)\), \((4,0)\) and \((0,6)\).
- Show that \(x^2+y^2+4x+2y-4=0\) and \(x^2+y^2-2x-6y-6=0\) cut orthogonally.