Circles, chords and tangents

What you’ll learn

• Find a circle’s centre and radius from its equation.
• Use circle geometry to find tangent and chord relationships.

Understand the idea

A circle consists of points a fixed distance from a centre. Squared distance gives its standard equation. To recover the centre and radius from an expanded equation, complete the square separately in $x$ and $y$, then collect the constants carefully.

A tangent is perpendicular to the radius at the contact point. The perpendicular from the centre to a chord meets the chord at its midpoint, allowing a right triangle to determine a chord length. An angle subtended by a diameter is a right angle, which can also be expressed using perpendicular gradients.

Key formulae

$$(x-a)^2+(y-b)^2=r^2$$

$$x^2+y^2+2gx+2fy+c=0\Rightarrow\text{centre }(-g,-f),\ r=\sqrt{g^2+f^2-c}$$

$$(x_1-a)(x-x_1)+(y_1-b)(y-y_1)=0\quad\text{(tangent at }(x_1,y_1)\text{)}$$

Worked example

Find the centre and radius of $x^2+y^2-6x+4y-12=0$ and the tangent at $(6,2)$.

1. Complete the squares: $(x-3)^2+(y+2)^2=25$.
2. The centre is $(3,-2)$ and radius is $5$. The radius to $(6,2)$ has gradient $4/3$.
3. The tangent has gradient $-3/4$ and passes through $(6,2)$.

Answer: Centre $(3,-2)$, radius $5$; tangent $3x+4y=26$.