Circle
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A circle is the set of all points in a plane that are a fixed distance from a given point in the plane. The given point is called the center of the circle which is denoted by $(h, k)$. The fixed distance is called the radius of the circle and is denoted by $r$.
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แกောแ်แါ applet แှ แกแှแ် $P(x, y)$ แို แွแ်แွဲแြแ့်แါ။
Standard Equation of a Circle
แေးแာแော แ แ်แိုแ်း၏ แแိုแแ် $O(h, k)$၊ แกแျแ်းแแ်แแ် $r$၊ แกแแ်းแေါ်แှိ แกแှแ်แแ ်แုแแ် $P(x, y)$ แြแ ်แแ် แိုแါแ ို့။แို့แြောแ့် $O(h, k)$ แှแ့် $P(x, y)$ แို့แแ် แกแျား $r$ unit แှိแော แျแ်းแแ ်แု၏ แกแ ွแ်းแှแ်แျား แြแ ်แแ်။
แို့แြောแ့် distance formula แกแ...
$\begin{array}{l} \sqrt{(x-h)^{2}+(y-k)^{2}}=r\\\\ \text { Squaring both sides, }\\\\ (x-h)^{2}+(y-k)^{2}=r^{2} \end{array}$
แို့แြောแ့် แกแျแ်းแแ် $r$ unit แှိแြီး แแို $(h, k)$ แှိแော แ แ်แိုแ်းแแ ်แု၏ standard equation แို
แုแြောแိုแ်แแ်။
General Equation of a Circle
G แกแျแ်းแแ် $r$ unit แှိแြီး แแို $(h, k)$ แှိแော แ แ်แိုแ်းแแ ်แု၏ standard equation แို แกแျแ်แြแ့်แါแ แ แ်แိုแ်းแแ ်แု၏ general equation แို แกောแ်แါแกแိုแ်း แแှိแါแแ်။$(x-h)^{2}+(y-k)^{2}=r^{2}$
$x^{2}-2 h x+h^{2}+y^{2}-2 k y+k^{2}=r^{2}$
$x^{2}+y^{2}-2 h x-2 k y+h^{2}+k^{2}-r^{2}=0$
Let $h^{2}+k^{2}-r^{2}=e$,
then the equation becomes
$x^{2}+y^{2}-2 h x-2 k y+e=0$
where centre $=(h, k)$.
Since $h^{2}+k^{2}-r^{2}=e$
$r^{2}=h^{2}+k^{2}-e$
$\therefore$ radius $=\sqrt{h^{2}+k^{2}-e}$
แို့แြောแ့် แ แ်แိုแ်းแแ ်แု၏ standard form equation แှแ့် general form equation แို့แို แกောแ်แါแแားแြแ့် แกแှแ ်แျုแ် แှแ်แားแိုแ်แါแแ်။
| Equation of a Circle | ||
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| Standard Form | General Form | |
| Equation | $(x-h)^{2}+(y-k)^{2}=r^{2}$ | $x^{2}+y^{2}-2 h x-2 k y+e=0$ |
| Centre | $(h, k)$ | $(h, k)$ |
| Radius | $r$ | $r=\sqrt{h^{2}+k^{2}-e}$ |
แแော့แျွแ် (Cone) แแ ်แုแို แြแ်แိုแ်แြแ်းแြောแ့် แြแ ်แာแော แแ်แแ့် แกแားแ ောแ်းแုံแ ံ (Conic Section) แแ ်แုแို แแို
| $Ax^2+Bxy+Cy^2+Dx+Ey+F = 0$ |
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แိုแော แျေแုแျ แုံแ ံแြแ့် แော်แြแိုแ် แြောแ်း Part (4) แွแ် แแ်แြแဲ့แြီး แြแ ်แแ်။
แ แ်แိုแ်းแแ ်แု၏ แီแျှแြแ်းแွแ် แกแแ်แွแ်แော်แြแဲ့แော แแားแกแ $A=C$, $B=0$, $D=-2h$, $E=-2k$ แှแ့် $F=e$ แြแ ်แแ်แု แှแ်แူแိုแ်แါแแ်။
Worked Examples
| Example (1) Write down the coordinates of the centre and the radius of each of thefollowing circles. (a) $x^{2}+y^{2}=81$ (b) $(x+4)^{2}+(y-6)^{2}=100$ (c) $x^{2}+\left(y+\displaystyle\frac{1}{3}\right)^{2}-16=0$ Solution $\begin{array}{ll} \textbf{(a)} & x^{2}+y^{2}=81 \\\\ &\text { Comparing with }(x-h)^{2}+(y-k)^{2}=r^{2}, \\\\ & h=k=0 \\\\ & r^{2}=81 \\\\ & r=9 \\\\ & \therefore \text { centre }=(0,0), \text { radius }=9\\\\ \textbf{(b)} & (x+4)^{2}+(y-6)^{2}=100\\\\ &\text { Comparing with }(x-h)^{2}+(y-k)^{2}=r^{2}, \\\\ & h=-4,\ k=6 \\\\ & r^{2}=100 \\\\ & r=10 \\\\ & \therefore \text { centre }=(-4,6), \text { radius }=10\\\\ \textbf{(c)} & x^{2}+\left(y+\displaystyle\frac{1}{3}\right)^{2}-16=0 \\\\ &\therefore\ x^{2}+\left(y+\displaystyle\frac{1}{3}\right)^{2}=16 \\\\ &\text { Comparing with }(x-h)^{2}+(y-k)^{2}=r^{2}, \\\\ & h=0, \ k=-\displaystyle\frac{1}{3} \\\\ & r^{2}=16 \\\\ & r=4 \\\\ & \therefore \text { centre }=\left(0,\ -\displaystyle\frac{1}{3}\right), \text { radius }=4 \end{array}$ |
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| Example (2) Find the equation of a circle with diameter PQ, where the coordinates of $P$ and $Q$ are $(−2, 1)$ and $(4, 3)$ respectively. Hence graph the circle. Solution Let the centre of the circle br $O$. Therefore, $O$ is the midpoint of circle. $\begin{aligned} \therefore \text { center } &=\left(\displaystyle\frac{-2+4}{2}, \displaystyle\frac{1+3}{2}\right) \\\\ &=\left(\displaystyle\frac{2}{2}, \displaystyle\frac{4}{2}\right) \\\\ &=(1,2) \\\\ \text { radius } &=O Q \\\\ &=\sqrt{(4-1)^{2}+(3-2)^{2}} \\\\ &=\sqrt{10} \end{aligned}$ $\therefore$ The equation of circle is $(x-1)^{2}+(y-2)^{2}=10$.  |
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| Example (3) Find the equation of a circle which touches the $x$-axis and whose centre is $O(−3, 5)$. Solution $\text{center} = (−3, 5)$ Since the circle touches $x$-axis, the point $(−3, 0)$ lies on the circumference of the circle. $\text{radius} = |5-0| = 5$. $\therefore$ The equation of circle is $(x+3)^{2}+(y-5)^{2}=5^2$. $(x+3)^{2}+(y-5)^{2}=25$. |
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| Example (4) Find the centre and the radius of the circle $x^2 + y^2 - 6x + 8y + 9 = 0$ Solution Method (1) Circle equation : $x^2 + y^2 - 6x + 8y + 9 = 0$ Comparing with $x^2 + y^2 - 2hx -2k y + e = 0$, $2h = 6,\ 2k = -8$ and $e=9$. $\therefore\ \ h=3, \ k=-4,\ e=9$. $\text{center} = (h,\ k) = (3,\ -4)$ $\begin{array}{l}\ \ \ \ \text{radius}=\sqrt{{{{h}^{2}}+{{k}^{2}}-e}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sqrt{{{{3}^{2}}+{{{(-4)}}^{2}}-9}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sqrt{{9+16-9}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4\end{array}$ Method (2) Circle equation : $x^2 + y^2 - 6x + 8y + 9 = 0$ Rearranging, we have $x^2 - 6x + y^2 + 8y = -9$ $x^2 - 2(3)x + 3^2 + y^2 + 2(4)y + 4^2 = -9+ 3^2 + 4^2$ $\therefore\ \ (x-3)^2 + (y + 4)^2 = 4^2$ Comparing with $(x-h)^2 + (y -k)^2 = r^2$, $\therefore\ \ h=3, \ k=-4,\ r=4$. $\text{center} = (h,\ k) = (3,\ -4)$ $\text{radius} = r = 4$ |
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Exercises
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