Equation of a circleAQA A-Level Maths: Revision notes

A circle is the set of points a fixed distance (the radius $r$) from a fixed point (the centre). If the centre is $(a,b)$ and $P(x,y)$ is on the circle, the distance formula gives $\sqrt{(x-a)^2+(y-b)^2}=r$, so $(x-a)^2+(y-b)^2=r^2.$ For a circle with centre at the origin this is $x^2+y^2=r^2$. Example: $(x-3)^2+(y+2)^2=25$ has centre $(3,-2)$ and radius $5$.

Write down the centre $(a,b)$ and find $r^2$ using what is given. If the circle has centre $C(2,-1)$ and passes through $P(5,3)$, then $r^2=CP^2=(5-2)^2+(3+1)^2=25$, giving $(x-2)^2+(y+1)^2=25$. If $AB$ is a diameter, the centre is the midpoint of $AB$ and the radius is half the length of $AB$. Leave $r^2$ as it is: you do not need to take a square root to write the equation.

A circle may be given as $x^2+y^2+2fx+2gy+c=0$. Complete the square for $x$ and $y$ separately: $x^2+2fx=(x+f)^2-f^2$. For $x^2+y^2-6x+10y+18=0$: $(x-3)^2-9+(y+5)^2-25+18=0$, so $(x-3)^2+(y+5)^2=16$. The centre is $(3,-5)$ and $r=4$. In general the centre is $(-f,-g)$ and $r^2=f^2+g^2-c$. If $r^2\le0$ the equation does not describe a circle.

For the circle $(x-a)^2+(y-b)^2=r^2$ and a point $(p,q)$ compare $(p-a)^2+(q-b)^2$ with $r^2$. If it equals $r^2$ the point is on the circle, if it is less the point is inside, and if it is greater the point is outside. For $(x-3)^2+(y+2)^2=25$: $(6,2)$ gives $9+16=25$ (on); $(3,2)$ gives $16$ (inside); $(-3,-2)$ gives $36$ (outside).

To find where a circle meets a line, substitute the line into the circle equation and solve the quadratic. The discriminant tells you the number of intersections: $b^2-4ac>0$ gives two points, $=0$ gives one (a tangent) and $<0$ gives none. Axes: put $x=0$ or $y=0$. Example: $x^2+y^2-6x+10y+18=0$ with $y=0$ gives $x^2-6x+18=0$ with discriminant $36-72<0$, so the circle does not meet the $x$-axis. Another example: $x^2+y^2-6x+2y-10=0$ with $y=x-6$ gives $x^2-8x+7=0$, so $x=1$ or $x=7$.