Distance and Midpoint Formulas and the Equation of a Circle

Q: How do I find the radius from a circle's equation?

Write it as (x - h)^2 + (y - k)^2 = r^2, then take the square root of the right-hand side.

How Pythagoras gives the distance between two points, how to find a midpoint, and how the distance formula becomes the equation of a circle, including finding the centre and radius by completing the square.

3 min read Supports 4 lessons

Distance between two points

The horizontal and vertical gaps between two points are the legs of a right-angled triangle. The straight-line distance between them is its hypotenuse, so Pythagoras gives it straight away.

Gaps of $6$ across and $8$ up, so the distance is $\sqrt{36 + 64} = 10$ .

Distance formula

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Tip

The order of subtraction does not matter, because each gap is squared. $(3 - (-3))^2$ and $(-3 - 3)^2$ are both $36$ .

The midpoint

The point halfway between two others sits halfway across and halfway up. Average the $x$ -coordinates and average the $y$ -coordinates.

Midpoint

M = \left(\frac{x_1 + x_2}{2},\; \frac{y_1 + y_2}{2}\right)

For $(-3, -2)$ and $(3, 6)$ : $M = (0, 2)$ .

The equation of a circle

A circle is every point at distance $r$ from its centre $(h, k)$ . Write that sentence with the distance formula and square both sides: the circle's equation appears.

$\text{distance from } (x, y) \text{ to } (h, k) = r$
The definition of a circle.

1/3

$(x - 2)^2 + (y - 1)^2 = 9$ : centre $(2, 1)$ , radius $3$ .

Watch out

$(x + 3)^2 + (y - 5)^2 = 16$ has centre $(-3, 5)$ and radius $4$ , not $16$ . The signs flip, and the right-hand side is $r^2$ .

Expanded equations

Sometimes the equation arrives multiplied out. Complete the square in $x$ and in $y$ to recover the centre and radius.

Worked example

Find the centre and radius of $x^2 + y^2 - 6x + 4y - 3 = 0$ .

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Group the $x$ and $y$ terms and complete each square.
$(x^2 - 6x + 9) + (y^2 + 4y + 4) = 3 + 9 + 4$

Key idea

In calculus you will find the slope of a circle at any point using implicit differentiation, starting from exactly this equation.

Common questions

How is the distance formula related to Pythagoras?: The horizontal and vertical gaps between two points form the legs of a right-angled triangle, and the distance is the hypotenuse, so $d^2 = (\Delta x)^2 + (\Delta y)^2$ .
How do I find the radius from a circle's equation?: Write it as $(x - h)^2 + (y - k)^2 = r^2$ , then take the square root of the right-hand side.

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Common questions

How is the distance formula related to Pythagoras?

The horizontal and vertical gaps between two points form the legs of a right-angled triangle, and the distance is the hypotenuse, so $d^2 = (\Delta x)^2 + (\Delta y)^2$ .

How do I find the radius from a circle's equation?

Write it as $(x - h)^2 + (y - k)^2 = r^2$ , then take the square root of the right-hand side.