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
