Write an Equation for Circle Graphs

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Question 1 of 7

1. Question
Write the equation of the circle with center $(0, 0)$ $(0,0)$ shown on the number plane below.
- 1. $x^{2} + y^{2} = 4$ $x^2+y^2=4$
- 2. $x^{2} + y^{2} = 2$ $x^2+y^2=2$
- 3. $x^{2} - y^{2} = 2$ $x^2-y^2=2$
- 4. $x^{2} - y^{2} = 4$ $x^2-y^2=4$
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Standard Circle Equation (Centre Origin)

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

Where the centre of the circle is $(0, 0)$ $(0,0)$ and the radius of the circle is $r$ $r$ .

Identify the radius of the circle from the points on its circumference.

The points $(2, 0)$ $(2,0)$ , $(0, 2)$ $(0,2)$ , $(- 2, 0)$ $(-2,0)$ and $(0, - 2)$ $(0,-2)$ all lie at a distance of $2$ $2$ from the centre $(0, 0)$ $(0,0)$ .

So the radius of the circle must be $2$ $2$

Substitute $r = 2$ $r=2$ into the standard equation given above, and then simplify the equation.

$x^{2} + y^{2} =$ $x^2+y^2=$ $2^{2}$ $2^2$

$x^{2} + y^{2} = 4$ $x^2+y^2=4$
Question 2 of 7

2. Question
Write the equation of the circle with center $(0, 0)$ $(0,0)$ shown on the number plane below.
- 1. $x^{2} - y^{2} = 10$ $x^2-y^2=10$
- 2. $x^{2} + y^{2} = 10$ $x^2+y^2=10$
- 3. $x^{2} + y^{2} = 100$ $x^2+y^2=100$
- 4. $x^{2} - y^{2} = 100$ $x^2-y^2=100$
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Standard Circle Equation (Centre Origin)

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

Where the centre of the circle is $(0, 0)$ $(0,0)$ and the radius of the circle is $r$ $r$ .

Identify the radius of the circle from the points on its circumference.

The points $(10, 0)$ $(10,0)$ , $(0, 10)$ $(0,10)$ , $(- 10, 0)$ $(-10,0)$ and $(0, - 10)$ $(0,-10)$ all lie at a distance of $10$ $10$ from the centre $(0, 0)$ $(0,0)$ .

So the radius of the circle must be $10$ $10$

Substitute $r = 10$ $r=10$ into the standard equation given above, and then simplify the equation.

$x^{2} + y^{2} =$ $x^2+y^2=$ $10^{2}$ $10^2$

$x^{2} + y^{2} = 100$ $x^2+y^2=100$
Question 3 of 7

3. Question
Write the equation of the circle with center $(0, 0)$ $(0,0)$ shown on the number plane below.
- 1. $x^{2} + y^{2} = 49$ $x^2+y^2=49$
- 2. $x^{2} - y^{2} = 7$ $x^2-y^2=7$
- 3. $x^{2} - y^{2} = 49$ $x^2-y^2=49$
- 4. $x^{2} + y^{2} = 7$ $x^2+y^2=7$
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Standard Circle Equation (Centre Origin)

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

Where the centre of the circle is $(0, 0)$ $(0,0)$ and the radius of the circle is $r$ $r$ .

Identify the radius of the circle from the points on its circumference.

The points $(7, 0)$ $(7,0)$ , $(0, 7)$ $(0,7)$ , $(- 7, 0)$ $(-7,0)$ and $(0, - 7)$ $(0,-7)$ all lie at a distance of $7$ $7$ from the centre $(0, 0)$ $(0,0)$ .

So the radius of the circle must be $7$ $7$

Substitute $r = 7$ $r=7$ into the standard equation given above, and then simplify the equation.

$x^{2} + y^{2} =$ $x^2+y^2=$ $7^{2}$ $7^2$

$x^{2} + y^{2} = 49$ $x^2+y^2=49$
Question 4 of 7

4. Question
Write the equation of the circle shown on the number plane below.
- 1. ${(x - 7)}^{2} + {(y - 7)}^{2} = 7$ $(x-7)^2+(y-7)^2=7$
- 2. $(x-7)^2+(y-7)^2=49$
- 3. $x^2+y^2=49$
- 4. $(x+7)^2+(y+7)^2=49$
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Standard Circle Equation

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

Where the centre of the circle is $($ $h$ $,$ $k$ $)$ and the radius of the circle is $r$ .

Identify the centre and radius of the circle from the points on its circumference.

The circle touches the $x$ -axis at $(7,0)$ and the $y$ -axis at $(0,7)$ , so the centre of the circle will be at $($ $7$ $,$ $7$ $)$ .

The points $(7,0)$ and $(0,7)$ lie at a distance of $7$ from the centre $($ $7$ $,$ $7$ $)$ .

So the radius of the circle must be $7$ .

Substitute $r=7$ , $h=7$ and $k=7$ into the standard equation given above, then simplify the equation.

$(x-$ $7$ $)^2+(y-$ $7$ $)^2=$ $7^2$

$(x-7)^2+(y-7)^2=49$
Question 5 of 7

5. Question
Write the equation of the circle shown on the number plane below.

The radius of the circle is $\sqrt{29}$ , the $x$ -coordinate of the centre is $8$ , and the point $(12,10)$ is on the circle circumference.
- 1. $x^2+y^2=29$
- 2. $(x-7)^2+(y-8)^2=\sqrt{29}$
- 3. $(x-8)^2+(y-8)^2=29$
- 4. $(x-7)^2+(y-8)^2=29$
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Standard Circle Equation

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

Where the centre of the circle is $($ $h$ $,$ $k$ $)$ and the radius of the circle is $r$ .

Substitute the information already known into the equation and simplify.

We know from the information given in the question that $r=\sqrt{29}$ and $k=8$

$(x-$ $h$ $)^2+(y-$ $8$ $)^2=$ $(\sqrt{29})^2$

$(x-$ $h$ $)^2+(y-$ $8$ $)^2=$ $29$

The point $($ $12$ $,$ $10$ $)$ lies on the circle. So substitute $x=12$ and $y=10$ into the equation and solve to find $h$ .

$($ $12$ $-$ $h$ $)^2+($ $10$ $-8)^2=29$

$(12-$ $h$ $)^2+2^2=29$

$(12-$ $h$ $)^2=25$

$\sqrt{(12-h)^2}=\pm \sqrt{25}$

$\sqrt{(12-h)^2}=\pm \sqrt{25}$

$12-h=\pm 5$

$h=7$ or $h=17$

From the diagram we can determine that $h=7$ .

Substitute $h=7$ into the circle equation found above.

$(x-$ $7$ $)^2+(y-$ $8$ $)^2=$ $29$

$(x-7)^2+(y-8)^2=29$
Question 6 of 7

6. Question
Write the equation of the circle shown on the number plane below.

The centre of the circle is $(4,0)$ and
the circle crosses the $x$ -axis at $(-3,0)$ .
- 1. $x^2+(y-4)^2=5$
- 2. $x^2+(y+5)^2=16$
- 3. $(x-4)^2+y^2=25$
- 4. $x^2+(y-4)^2=25$
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Standard Circle Equation

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

Where the centre of the circle is $($ $h$ $,$ $k$ $)$ and the radius of the circle is $r$ .

Substitute the information already known into the equation and simplify.

We know from the information given in the question that $h=0$ and $k=4$ .

$(x-$ $0$ $)^2+(y-$ $4$ $)^2=$ $r^2$

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

The point $($ $-3$ $,$ $0$ $)$ lies on the circle. So substitute $x=-3$ and $y=0$ into the equation and solve to find $r$ .

$($ $-3$ $)^2+($ $0$ $-4)^2=r^2$

$9+16=r^2$

$25=r^2$

$\sqrt{25}=\sqrt{r^2}$

$r=5$ .

Substitute $r=5$ into the circle equation found above and simplify.

$x^2+(y-$ $4$ $)^2=$ $5^2$

$x^2+(y-4)^2=25$
Question 7 of 7

7. Question
Write the equation of the circle shown on the number plane below.

The circle has an $x$ -intercept of $-6$ , a $y$ -intercept of $4$ and passes through the origin $(0,0)$ .
- 1. $(x+3)^2+(y-2)^2=13$
- 2. $(x+3)^2+(y-2)^2=\sqrt{13}$
- 3. $(x+3)^2+(y-2)^2=52$
- 4. $(x-3)^2+(y+2)^2=13$
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Standard Circle Equation

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

Where the centre of the circle is $($ $h$ $,$ $k$ $)$ and the radius of the circle is $r$ .

Draw a line between the $x$ and $y$ intercepts of the circle to create a diameter of the circle.

Use Pythagoras’ Theorem to find the length of the diameter, then halve this to find the radius

$d^2=6^2+4^2$

$d^2=52$

$d=\sqrt{52}$

The radius is half this: $\frac{d}{2}=\frac{\sqrt{52}}{2}\frac{\sqrt{52}}{\sqrt{4}}=\sqrt{13}$

$r=\sqrt{13}$ .

Find the halfway point along the diameter to give the center of the circle.

The $x$ -coordinate is given by $\frac{-6}{2}=-3$

The $y$ -coordinate is given by $\frac{4}{2}=2$

The centre of the circle is $(-3,2)$

$h=-3$ and $k=2$

Substitute $r=\sqrt{13}$ , $h=-3$ and $k=2$ into the standard equation given above, then simplify the equation.

$(x-$ $(-3)$ $)^2+(y-$ $2$ $)^2=$ $\sqrt{13}^2$

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

Write an Equation for Circle Graphs

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Quiz summary

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1. Question

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Well Done!

Standard Circle Equation (Centre Origin)

2. Question

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Nice Job!

Standard Circle Equation (Centre Origin)

3. Question

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Excellent!

Standard Circle Equation (Centre Origin)

4. Question

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Fantastic!

Standard Circle Equation

5. Question

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Keep Going!

Standard Circle Equation

6. Question

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Correct!

Standard Circle Equation

7. Question

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Great Work!

Standard Circle Equation

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