Hi Have found the one of the circle formulas to be x^2+y^2-4x-10y+19=0. My problem is finding the radius and center with this equation. Can you help me with this question. Also, please show the steps to how you got your answer. Thanks!
Hi Have found the one of the circle formulas to be x^2+y^2-4x-10y+19=0. My problem is finding the radius and center with this equation. Can you help me with this question. Also, please show the steps to how you got your answer. Thanks!
$\displaystyle x^2+y^2-4x-10y+19=0$
Rearrange
$\displaystyle x^2-4x+y^2-10y+19=0$
Now complete the square in $\displaystyle x$ and $\displaystyle y$
$\displaystyle (x^2-4x+4)-4+(y^2-10y+25)-25+19=0$
$\displaystyle (x-2)^2-4+(y-5)^2-25+19=0$
$\displaystyle (x-2)^2+(y-5)^2-29+19=0$
$\displaystyle (x-2)^2+(y-5)^2-10=0$
$\displaystyle (x-2)^2+(y-5)^2=10$
$\displaystyle (x-2)^2+(y-5)^2=(\sqrt{10})^2$
Can you see the radius and centre?
Hi starrynight,
there are 2 general ways, one way being to find "g" and "f" by inspection from
$\displaystyle x^2+y^2+2gx+2fy+c=0$
then the centre is $\displaystyle (-g,-f)$
then use g and f to find r, using $\displaystyle r=\sqrt{g^2+f^2-c}$
$\displaystyle 2g=-4,\ g=-2,\ -g=2$
$\displaystyle 2f=-10,\ f=-5,\ -f=5$
$\displaystyle g^2=4,\ f^2=25$
$\displaystyle c=19$
centre is (2,5)
$\displaystyle r=\sqrt{4+25-19}=\sqrt{10}$
Alternatively, complete the squares for x and y, using half the coefficents
of x and y.
$\displaystyle x^2-4x\color{blue}+(-2)^2\color{black}+y^2-10y\color{blue}+(-5)^2\color{black}+19\color{blue}-(-2)^2-(-5)^2=0$
$\displaystyle x^2-4x+4+y^2-10y+25+19-4-25=0$
$\displaystyle (x-2)^2+(y-5)^2-10=0$
$\displaystyle (x-2)^+(y-5)^2=(\sqrt{10})^2=r^2$
centre is (2,5), radius is $\displaystyle \sqrt{10}$