Graph Circles in Expanded Form

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Question 1 of 2

1. Question
Identify the centre and radius, then sketch the following circle.

$2 x^{2} + 4 x + 2 y^{2} - 16 y + 18 = 0$ $2x^2+4x+2y^2-16y+18=0$
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In order to graph the circle, first find the centre and radius

The centre and radius can be read from the standard form equation

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

where the centre is $($ $($ $h$ $h$ $,$ $,$ $k$ $k$ $)$ $)$ and the radius is $r$ $r$

Rewrite the given equation in standard form.

Start by dividing both sides of the equation by 2.

$2 x^{2} + 4 x + 2 y^{2} - 16 y + 18 = 0$ $2x^2+4x+2y^2-16y+18=0$

$x^{2} + 2 x$ $x^2+2x$ $+$ $+$ $y^{2} - 8 y$ $y^2-8y$ $+ 9 = 0$ $+9=0$

Now complete the squares.

Start by adding constants to the expressions $x^{2} + 2 x$ $x^2+2x$ and $y^{2} - 8 y$ $y^2-8y$ .

The constant to add to $x^{2} + 2 x$ $x^2+2x$ is half the coefficient of $x$ $x$ squared.

${(\frac{1}{2} \times 2)}^{2} = 1^{2} =$ $(\frac{1}{2}\times2)^2=1^2=$ $1$ $1$

The constant to add to $y^{2} - 8 y$ $y^2-8y$ is half the coefficient of $y$ $y$ squared.

${(\frac{1}{2} \times (- 8))}^{2} = 4^{2} =$ $(\frac{1}{2}\times(-8))^2=4^2=$ $16$ $16$

Since you are adding $1$ $1$ and $16$ $16$ to the left-hand side, add $1$ $1$ $+$ $+$ $16$ $16$ $= 17$ $=17$ to the right-hand side

$x^{2} + 2 x + 1$ $x^2+2x+1$ $+$ $+$ $y^{2} - 8 x + 16$ $y^2-8x+16$ $+ 9 = 17$ $+9=17$

Now use the facts that $x^{2} + 2 x + 1 = {(x + 1)}^{2}$ $x^2+2x+1=(x+1)^2$ and $y^{2} - 8 x + 16 = {(y - 4)}^{2}$ $y^2-8x+16=(y-4)^2$ to rewrite these parts in the equation.

${(x + 1)}^{2}$ $(x+1)^2$ $+$ $+$ ${(y - 4)}^{2}$ $(y-4)^2$ $+ 9 = 17$ $+9=17$

Subtract $9$ $9$ from both sides of the equation and finish rewriting in standard form.

${(x + 1)}^{2} + {(y - 4)}^{2} = 8$ $(x+1)^2+(y-4)^2=8$

$(x -$ $(x-$ $(- 1)$ $(-1)$ $)^{2} + (y -$ $)^2+(y-$ $4$ $4$ $)^{2} =$ $)^2=$ ${\sqrt{8}}^{2}$ $\sqrt{8}^2$

In this equation $h = - 1$ $h=-1$ $,$ $,$ $k = 4$ $k=4$ and $r = \sqrt{8}$ $r=\sqrt{8}$

The centre is $($ $($ $- 1$ $-1$ $,$ $,$ $4$ $4$ $)$ $)$ and the radius is $\sqrt{8} \approx 2.828$ $\sqrt{8}\approx 2.828$

To sketch the circle first plot the centre $($ $-1$ $,$ $4$ $)$ .

Since the radius is $2.828$ , plot the points which are $2.828$ units above, below, left and right of the centre.

Complete the sketch
Question 2 of 2

2. Question
Identify the centre and radius, then sketch the following circle.

$x^2+y^2-4x+6y-3=0$
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In order to graph the circle, first find the centre and radius

The centre and radius can be read from the standard form equation

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

where the centre is $($ $h$ $,$ $k$ $)$ and the radius is $r$

Rewrite the left-hand side so the terms with the same variables are together.

$x^2-4x$ $+$ $y^2+6y$ $-3=0$

Now complete the squares.

Start by adding constants to the expressions $x^2-4x$ and $y^2+6y$ .

The constant to add to $x^2-4x$ is half the coefficient of $x$ squared.

$(\frac{1}{2}\times (-4))^2=2^2=$ $4$

The constant to add to $y^2+6y$ is half the coefficient of $y$ squared.

$(\frac{1}{2}\times 6)^2=3^2=$ $9$

Since you are adding $4$ and $9$ to the left-hand side, add $4$ $+$ $9$ $=13$ to the right-hand side

$x^2-4x+4$ $+$ $y^2+6y+9$ $-3=13$

Now use the facts that $x^2-4x+4=(x-2)^2$ and $y^2+6x+9=(y+3)^2$ to rewrite these parts in the equation.

$(x-2)^2$ $+$ $(y+3)^2$ $-3=13$

Add $3$ to both sides of the equation and finish rewriting in standard form.

$(x-2)^2+(y+3)^2=16$

$(x-$ $2$ $)^2+(y-$ $(-3)$ $)^2=$ $4^2$

In this equation $h=2$ $,$ $k=-3$ and $r=4$

The centre is $($ $2$ $,$ $-3$ $)$ and the radius is $4$ .

To sketch the circle first plot the centre $($ $2$ $,$ $-3$ $)$ .

Since the radius is $4$ , plot the points which are $4$ units above, below, left and right of the centre.

Complete the sketch

Graph Circles in Expanded Form

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