Worked examples · Sheet 3

Circles and completing the square

Theory used: Circle Completing the square Distance between two points and midpoint

Exercise 1

Write the equation of the circle with center $C (2, - 1)$ and radius $r = 3$ , in standard form and in general form.

The standard form is

(x - 2)^{2} + (y + 1)^{2} = 9.

Expanding the squares:

x^{2} - 4 x + 4 + y^{2} + 2 y + 1 x^{2} + y^{2} - 4 x + 2 y - 4 = 9 = 0.

Result: $(x - 2)^{2} + (y + 1)^{2} = 9$ , that is $x^{2} + y^{2} - 4 x + 2 y - 4 = 0$ .

The standard form is written directly from the data: in place of $x_{0}$ and $y_{0}$ one puts the coordinates of the center and on the right the square of the radius. The sign inside the second square is plus, because the ordinate of the center is $- 1$ and $y - (- 1) = y + 1$ . The general form is obtained by expanding the squares with the special products and moving everything to the left. A check on the coefficients: in the general form the coefficient of $x$ is $- 2 x_{0} = - 4$ and that of $y$ is $- 2 y_{0} = 2$ , in agreement with the given center.

Exercise 2

Find the center and radius of the circle of equation

x^{2} + y^{2} - 6 x + 4 y - 3 = 0.

Complete the squares in the terms in $x$ and in the terms in $y$ :

x^{2} - 6 x y^{2} + 4 y = (x - 3)^{2} - 9, = (y + 2)^{2} - 4,

therefore the equation becomes

(x - 3)^{2} - 9 + (y + 2)^{2} - 4 - 3 = 0, (x - 3)^{2} + (y + 2)^{2} = 16.

Result: center $C (3, - 2)$ and radius $r = 4$ .

The equation is in general form, with $x^{2}$ and $y^{2}$ having coefficient $1$ and no $x y$ term, therefore it can represent a circle. Completing the square groups the terms in $x$ and those in $y$ : to $x^{2} - 6 x$ one adds and subtracts the square of half of $6$ , and likewise for $y^{2} + 4 y$ with half of $4$ . Moving the constants to the right gives the standard form, from which center and radius are read: the center is the point that makes both squares vanish and the radius is the square root of the constant term, $r = 16 = 4$ , positive as required.

Exercise 3

Find the center and radius of the circle of equation

2 x^{2} + 2 y^{2} - 4 x + 8 y - 8 = 0.

The coefficients of $x^{2}$ and $y^{2}$ are both equal to $2$ , so divide the equation by $2$ :

x^{2} + y^{2} - 2 x + 4 y - 4 = 0.

Completing the squares:

(x - 1)^{2} - 1 + (y + 2)^{2} - 4 - 4 (x - 1)^{2} + (y + 2)^{2} = 0 = 9.

Result: center $C (1, - 2)$ and radius $r = 3$ .

The general form of the circle has the squares with coefficient $1$ , so the first operation divides the whole equation by $2$ ; the division is legitimate because it does not change the solution set. From there the procedure is the same as in the previous exercise: one completes the square in $x$ with half of $2$ and the one in $y$ with half of $4$ , then collects the constants on the right. A check by substitution: the point $(4, - 2)$ , which is at distance $3$ from the center along the $x$ -axis, satisfies the original equation, since $2 \cdot 16 + 2 \cdot 4 - 16 - 16 - 8 = 32 + 8 - 40 = 0$ .