circles and radius and center

**starrynight** · April 23rd 2010, 01:59 PM

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!

**pickslides** · April 23rd 2010, 02:15 PM

$x^2+y^2-4x-10y+19=0$

Rearrange

$x^2-4x+y^2-10y+19=0$

Now complete the square in $x$ and $y$

$(x^2-4x+4)-4+(y^2-10y+25)-25+19=0$

$(x-2)^2-4+(y-5)^2-25+19=0$

$(x-2)^2+(y-5)^2-29+19=0$

$(x-2)^2+(y-5)^2-10=0$

$(x-2)^2+(y-5)^2=10$

$(x-2)^2+(y-5)^2=(\sqrt{10})^2$

Can you see the radius and centre?

**Archie Meade** · April 23rd 2010, 02:23 PM

Originally Posted by starrynight

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 starrynight,

there are 2 general ways, one way being to find "g" and "f" by inspection from

$x^2+y^2+2gx+2fy+c=0$

then the centre is $(-g,-f)$

then use g and f to find r, using $r=\sqrt{g^2+f^2-c}$

$2g=-4,\ g=-2,\ -g=2$

$2f=-10,\ f=-5,\ -f=5$

$g^2=4,\ f^2=25$

$c=19$

centre is (2,5)

$r=\sqrt{4+25-19}=\sqrt{10}$

Alternatively, complete the squares for x and y, using half the coefficents
of x and y.

$x^2-4x\color{blue}+(-2)^2\color{black}+y^2-10y\color{blue}+(-5)^2\color{black}+19\color{blue}-(-2)^2-(-5)^2=0$

$x^2-4x+4+y^2-10y+25+19-4-25=0$

$(x-2)^2+(y-5)^2-10=0$

$(x-2)^+(y-5)^2=(\sqrt{10})^2=r^2$

centre is (2,5), radius is $\sqrt{10}$

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