The equation of a circle

Covers AS91573

What a circle is

A circle is the set of all points that are a fixed distance from a fixed point.
The fixed point is the centre; the fixed distance is the radius $r$ .
A circle is a conic section — the curve made when a plane cuts a cone at right angles to its axis.

A circle centred at the origin with radius r

The equation of a circle at the origin

A circle centred at the origin $(0, 0)$ with radius $r$ has equation:

$x^2 + y^2 = r^2$

$x$ , $y$ — the coordinates of any point on the circle.
$r$ — the radius; note the equation uses $r^2$ , not $r$ .
Every point that satisfies the equation lies exactly $r$ from the centre — this comes straight from the distance formula.

The equation of a circle with centre $(h, k)$

Move the centre to $(h, k)$ and the equation becomes:

$(x - h)^2 + (y - k)^2 = r^2$

$h$ — the $x$ -coordinate of the centre.
$k$ — the $y$ -coordinate of the centre.
Read the signs carefully: a centre of $(3, -2)$ gives $(x - 3)^2 + (y + 2)^2 = r^2$ .

Reading off centre and radius

Match the given equation to the standard form.
The centre is $(h, k)$ — the numbers subtracted inside each bracket.
The radius is $r = \sqrt{\text{right-hand side}}$ — always take the square root.

Worked ExampleCentre and radius from standard form

A circle has equation $(x - 4)^2 + (y + 3)^2 = 36$ . State its centre and radius, and find whether the point $(10, -3)$ lies on it.

Step 1 — Match to the standard form

Compare with $(x - h)^2 + (y - k)^2 = r^2$ . The numbers subtracted give $h = 4$ and $k = -3$ .

$\text{Centre} = (4, -3)$

Step 2 — Take the square root for the radius

$r = \sqrt{36} = 6$

Step 3 — Test the point $(10, -3)$

Substitute $x = 10$ , $y = -3$ into the left-hand side:

$(10 - 4)^2 + (-3 + 3)^2 = 6^2 + 0^2 = 36$

This equals the right-hand side, so the point satisfies the equation.

$\boxed{\text{Centre } (4, -3), \; r = 6; \; (10, -3) \text{ lies on the circle}}$

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

Write down the centre and radius of the circle $(x + 2)^2 + (y - 5)^2 = 49$ .

Merit

A circle has centre $(3, -1)$ and passes through the point $(7, 2)$ . Find its equation in the form $(x - h)^2 + (y - k)^2 = r^2$ .

Excellence

The points $A(1, 2)$ and $B(7, 10)$ are the ends of a diameter of a circle. Find the equation of the circle, and justify each step.

The equation of a circle

Covers AS91573

What a circle is

A circle is the set of all points that are a fixed distance from a fixed point.
The fixed point is the centre; the fixed distance is the radius $r$ .
A circle is a conic section — the curve made when a plane cuts a cone at right angles to its axis.

A circle centred at the origin with radius r

The equation of a circle at the origin

A circle centred at the origin $(0, 0)$ with radius $r$ has equation:

$x^2 + y^2 = r^2$

$x$ , $y$ — the coordinates of any point on the circle.
$r$ — the radius; note the equation uses $r^2$ , not $r$ .
Every point that satisfies the equation lies exactly $r$ from the centre — this comes straight from the distance formula.

The equation of a circle with centre $(h, k)$

Move the centre to $(h, k)$ and the equation becomes:

$(x - h)^2 + (y - k)^2 = r^2$

$h$ — the $x$ -coordinate of the centre.
$k$ — the $y$ -coordinate of the centre.
Read the signs carefully: a centre of $(3, -2)$ gives $(x - 3)^2 + (y + 2)^2 = r^2$ .

Reading off centre and radius

Match the given equation to the standard form.
The centre is $(h, k)$ — the numbers subtracted inside each bracket.
The radius is $r = \sqrt{\text{right-hand side}}$ — always take the square root.

Worked ExampleCentre and radius from standard form

A circle has equation $(x - 4)^2 + (y + 3)^2 = 36$ . State its centre and radius, and find whether the point $(10, -3)$ lies on it.

Step 1 — Match to the standard form

Compare with $(x - h)^2 + (y - k)^2 = r^2$ . The numbers subtracted give $h = 4$ and $k = -3$ .

$\text{Centre} = (4, -3)$

Step 2 — Take the square root for the radius

$r = \sqrt{36} = 6$

Step 3 — Test the point $(10, -3)$

Substitute $x = 10$ , $y = -3$ into the left-hand side:

$(10 - 4)^2 + (-3 + 3)^2 = 6^2 + 0^2 = 36$

This equals the right-hand side, so the point satisfies the equation.

$\boxed{\text{Centre } (4, -3), \; r = 6; \; (10, -3) \text{ lies on the circle}}$

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

Write down the centre and radius of the circle $(x + 2)^2 + (y - 5)^2 = 49$ .

Merit

A circle has centre $(3, -1)$ and passes through the point $(7, 2)$ . Find its equation in the form $(x - h)^2 + (y - k)^2 = r^2$ .

Excellence

The points $A(1, 2)$ and $B(7, 10)$ are the ends of a diameter of a circle. Find the equation of the circle, and justify each step.

The equation of a circle

What a circle is

The equation of a circle at the origin

The equation of a circle with centre (h,k)(h, k)(h,k)

Reading off centre and radius

Step 1 — Match to the standard form

Step 2 — Take the square root for the radius

Step 3 — Test the point (10,−3)(10, -3)(10,−3)

Test yourself

Practice by grade

The equation of a circle

What a circle is

The equation of a circle at the origin

The equation of a circle with centre (h,k)(h, k)(h,k)

Reading off centre and radius

Step 1 — Match to the standard form

Step 2 — Take the square root for the radius

Step 3 — Test the point (10,−3)(10, -3)(10,−3)

Test yourself

Practice by grade

The equation of a circle with centre $(h, k)$

Step 3 — Test the point $(10, -3)$

The equation of a circle with centre $(h, k)$

Step 3 — Test the point $(10, -3)$