Write an Equation from Circle Graphs

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Question 1 of 7

1. Question
Write the equation of the circle with center $(0,0)$ shown on the number plane below.
- 1. $x^2-y^2=4$
- 2. $x^2+y^2=2$
- 3. $x^2-y^2=2$
- 4. $x^2+y^2=4$
Hint

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Standard Circle Equation (Centre Origin)

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

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

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

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

So the radius of the circle must be $2$

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

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

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

2. Question
Write the equation of the circle with center $(0,0)$ shown on the number plane below.
- 1. $x^2-y^2=10$
- 2. $x^2-y^2=100$
- 3. $x^2+y^2=10$
- 4. $x^2+y^2=100$
Hint

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Nice Job!

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Standard Circle Equation (Centre Origin)

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

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

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

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

So the radius of the circle must be $10$

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

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

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

3. Question
Write the equation of the circle with center $(0,0)$ shown on the number plane below.
- 1. $x^2+y^2=7$
- 2. $x^2+y^2=49$
- 3. $x^2-y^2=49$
- 4. $x^2-y^2=7$
Hint

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Correct

Excellent!

Incorrect

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Standard Circle Equation (Centre Origin)

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

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

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

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

So the radius of the circle must be $7$

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

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

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

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

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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 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. $(x-7)^2+(y-8)^2=29$
- 2. $(x-8)^2+(y-8)^2=29$
- 3. $(x-7)^2+(y-8)^2=\sqrt{29}$
- 4. $x^2+y^2=29$
Hint

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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. $x^2+(y-4)^2=25$
- 2. $(x-4)^2+y^2=25$
- 3. $x^2+(y+5)^2=16$
- 4. $x^2+(y-4)^2=5$
Hint

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Correct

Correct!

Incorrect

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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. $(x+3)^2+(y-2)^2=52$
- 3. $(x-3)^2+(y+2)^2=13$
- 4. $(x+3)^2+(y-2)^2=\sqrt{13}$
Hint

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Correct

Well Done

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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’s Theorem to find the length of the diameter, then half 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 from Circle Graphs

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Quiz summary

Information

1. Question

Hint

Well Done!

Standard Circle Equation (Centre Origin)

2. Question

Hint

Nice Job!

Standard Circle Equation (Centre Origin)

3. Question

Hint

Excellent!

Standard Circle Equation (Centre Origin)

4. Question

Hint

Fantastic!

Standard Circle Equation

5. Question

Hint

Keep Going!

Standard Circle Equation

6. Question

Hint

Correct!

Standard Circle Equation

7. Question

Hint

Great Work!

Standard Circle Equation

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