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Finding the Equation of a Circle Graph>
Finding the Equation of a Circle GraphFinding the Equation of a Circle Graph
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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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Well Done!
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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
Help VideoCorrect
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
Help VideoCorrect
Excellent!
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 `(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
Help VideoCorrect
Fantastic!
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`.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.
Hint
Help VideoCorrect
Keep Going!
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 `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)`.
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+(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)`.
Hint
Help VideoCorrect
Great Work!
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`.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`