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The general form of the equation of a circle is x2+y2−4x−8y−5=0.

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The general form of the equation of a circle is x2+y2−4x−8y−5=0.

What are the coordinates of the center of the circle?

jbouyer  May 16, 2017
#1
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x^2 + y^2 − 4x − 8y − 5 = 0      complete the square on x and y

( x^2 - 4x  + 4)  + ( y^2 - 8y + 16)  - 5 - 4 - 16  = 0   simplify

(x - 2)^2  +  ( y - 4)^2  -  25  = 0       add 25 to both sides

(x - 2)^2  +  ( y - 4)^2   =  25

The center  is  ( 2, 4)

CPhill  May 16, 2017
#2
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The general form of the equation of a circle is x2+y2−4x−8y−5=0.

What are the coordinates of the center of the circle?

A circle can be defined as the locus of all points that satisfy the equation

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

where r is the radius of the circle,
and h,k are the coordinates of its center.

The general Form is:

$$x^2+y^2 +ax+by+c = 0$$

Standard Form to general Form:

$$\begin{array}{|rcll|} \hline (x-h)^2 + (y-k)^2 &=& r^2 \\ x^2-2xh+h^2+y^2-2yk+k^2 &=& r^2 \\ x^2+y^2+x\cdot\underbrace{(-2h)}_{=a}+y\cdot\underbrace{(-2k)}_{=b}+\underbrace{h^2+k^2-r^2}_{=c} &=& 0 \\ \hline \end{array}$$

h,k and r ?

$$\begin{array}{|rcll|} \hline x^2+y^2+x\cdot\underbrace{(-2h)}_{=a}+y\cdot\underbrace{(-2k)}_{=b}+\underbrace{h^2+k^2-r^2}_{=c} &=& 0 \\\\ a &=&-2h\\ \color{red}h &\color{red}=& \color{red}-\frac{a}{2} \\\\ b &=&-2k\\ \color{red}k &\color{red}=& \color{red}-\frac{b}{2} \\\\ c &=&h^2+k^2-r^2\\ c &=&(-\frac{a}{2})^2+(-\frac{b}{2})^2-r^2\\ c &=& \frac{a^2+b^2}{4} -r^2\\ r^2 &=& \frac{a^2+b^2}{4} -c \\ \color{red}r &\color{red}=& \color{red} \sqrt{\frac{a^2+b^2}{4} -c} \\ \hline \end{array}$$

If we have a,b and c, we can calculate h,k and r:

$$\begin{array}{|lcll|} \hline \mathbf{x^2+y^2 +ax+by+c = 0} \\ h = -\dfrac{a}{2} \\ k = -\dfrac{b}{2} \\ r = \sqrt{\dfrac{a^2+b^2}{4} -c} \\ \hline \end{array}$$

$$a=-4\\ b=-8\\ c=-5$$

$$\begin{array}{|lcll|} \hline \mathbf{x^2+y^2 -4x-8y-5 = 0} \\\\ h = -\dfrac{-4}{2} \\ \mathbf{h = 2} \\\\ k = -\dfrac{-8}{2} \\ \mathbf{k = 4} \\\\ r = \sqrt{\dfrac{(-4)^2+(-8)^2}{4} -(-5)} \\ r = \sqrt{\dfrac{16+64}{4} +5} \\ r = \sqrt{20 +5} \\ r = \sqrt{25} \\ \mathbf{r = 5} \\ \hline \end{array}$$

The coordinates of the center of the circle is (2,4) and the radius is 5

heureka  May 16, 2017

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