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Write an Equation for Circle GraphsWrite 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)` shown on the number plane below.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 at 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.Hint
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Nice Job!
Incorrect
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 at 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.Hint
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Excellent!
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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 at 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.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 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``(x7)^2+(y7)^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.Hint
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Keep Going!
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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{(12h)^2}=\pm \sqrt{25}``\sqrt{(12h)^2}=\pm \sqrt{25}``12h=\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``(x7)^2+(y8)^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)`.Hint
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Correct!
Incorrect
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+(y4)^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+(y4)^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)`.Hint
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Great Work!
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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+(y2)^2=13`
Quizzes
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 Graph Parabolas Given the Vertex, Focus and Directrix 2
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