The equation of a circle

A point $\displaystyle (x,y)$ is on the circle with centre $\displaystyle (a,b)$ and radius $\displaystyle R$ exactly when its distance from the centre is $\displaystyle R$:

We square both sides and get the general equation of a circle:

When the centre is the origin, $\displaystyle {a}={0}$ and $\displaystyle {b}={0}$, and we get the canonical equation:

The centre and the radius are read from the general equation. Mind the signs: the equation is written with minus signs, $\displaystyle (x-a)$ and $\displaystyle (y-b)$.

• $\displaystyle {(x+1)^{2}+(y-3)^{2}}={25}$: here $\displaystyle {x+1}={x-(-1)}$, so the centre is $\displaystyle (-1,3)$ and the radius $\displaystyle {\sqrt{25}}={5}$.
• $\displaystyle {x^{2}+y^{2}}={4}$: the centre is $\displaystyle (0,0)$ and the radius $\displaystyle 2$ (the dashed one).
• $\displaystyle {x^{2}+(y-4)^{2}}={7}$: the centre is $\displaystyle (0,4)$ and the radius $\displaystyle \sqrt{7}$.

A point is on the circle if its coordinates satisfy the equation. Substitute them and compare with $\displaystyle R^{2}$: equal – on the circle; less – inside it; more – outside it.

Example: the circle $\displaystyle {(x-2)^{2}+(y-5)^{2}}={25}$ and the point $\displaystyle A(6,2)$.

The result equals $\displaystyle 25$, so $\displaystyle A$ is on the circle.

When the centre and a point on the circle are known, the radius is the distance between them. Usually no root is needed: the equation needs $\displaystyle R^{2}$.

Example: the centre of the circle is $\displaystyle (3,-2)$, and the circle passes through $\displaystyle (6,2)$.

The equation of the circle: $\displaystyle {(x-3)^{2}+(y+2)^{2}}={25}$.

The centre of a circle is the midpoint of every diameter, and the radius is half the diameter.

Example: $\displaystyle AB$ is a diameter, $\displaystyle A(6,2)$ and $\displaystyle B(-2,8)$.

• The centre is the midpoint of $\displaystyle AB$: $\displaystyle {\left(\frac{6+(-2)}{2},\frac{2+8}{2}\right)}={(2,5)}$.
• The diameter: $\displaystyle {AB}={\sqrt{(6+2)^{2}+(2-8)^{2}}}={\sqrt{100}}={10}$, so $\displaystyle {R}={5}$.
• The equation of the circle: $\displaystyle {(x-2)^{2}+(y-5)^{2}}={25}$.

Find the centre and the radius of each circle.

$\displaystyle {(x-4)^{2}+(y+1)^{2}}={16}$

$\displaystyle {(x+5)^{2}+y^{2}}={3}$

$\displaystyle {x^{2}+y^{2}}={49}$

Write the equation of the circle in the drawing. Its centre and radius are whole numbers.

The circle: $\displaystyle {(x-2)^{2}+(y+3)^{2}}={25}$.

Which of the points is on the circle, and which is inside it? Explain.

The centre of the circle is $\displaystyle M(-3,2)$, and the circle passes through the point $\displaystyle P(1,5)$.
Write the equation of the circle.

$\displaystyle AB$ is a diameter of a circle: $\displaystyle A(-1,4)$ and $\displaystyle B(5,-4)$.
Write the equation of the circle.