A circle has the equation x^{2}+y^{2}-6x+12y+9=0. What is the di

Step 1
The equation of the circle is given as,
${\displaystyle x^2+y^2-6x+12y+9=0}$
The general equation of the circle with center at ${\displaystyle (h,k)}$ and radius r is given as,
${\displaystyle {(x-h)}^2+{(y-k)}^2=r^2}$
On simplifying the equation of the circle given as,
${\displaystyle x^2+y^2-6x+12y+9=0}$
${\displaystyle (x^2-6x)+(y^2+12y)+9=0}$
${\displaystyle (x^2-2\times x\times 3+3^2-3^2)+(y^2+2\times y\times 6+6^2-6^2)+9=0}$
${\displaystyle (x^2-2\times x\times 3+3^2)+(y^2+2\times y\times 6+6^2)+9-3^2-6^2=0}$
${\displaystyle {(x-3)}^2+{(y+6)}^2-36=0}$
${\displaystyle {(x-3)}^2+{(y-(-6))}^2={\left(6\right)}^2}$
On comparing the equations, we get the center of the circle as ${\displaystyle (3,-6)}$ and the radius of the circle as 6 units.
Step 2
The equation of the line from which the distance between center and the line is to be calculated is,
${\displaystyle y=-2x+10}$
The general equation of the line is,
${\displaystyle y=-2x+10}$
${\displaystyle 2x+y-10=0}$
The distance of line ${\displaystyle ax+by+c=0}$ from the point ${\displaystyle (x_1,y_1)}$ is given as,
${\displaystyle d=\frac{a{x}_1+b{y}_1+c}{\left|\sqrt{a^2+b^2}\right|}}$
Putting the center and the equation of the line, we get
${\displaystyle d=\left|\frac{2\times \left(3\right)+(-6)-10}{\sqrt{{\left(2\right)}^2+{\left(1\right)}^2}}\right|}$
${\displaystyle =\left|\frac{-10}{\sqrt{5}}\right|}$
${\displaystyle =2\sqrt{5}}$
Therefore, the required distance between the center of the circle and the given line is ${\displaystyle 2\sqrt{5}}$ units.

Step 1
Equation of circle can be written as
$(x-3{)}^2+(y+6{)}^2=36$
Center of circle
$=(3,-6)$
Shortest distance
$=d=|\frac{-6-6-10}{\sqrt{2^2+1^2}}|=\frac{22}{\sqrt{5}}\sim 9.8387units$

The shortest distance to the line from any point is a segment of a perpendicular line that passes through the point.
$h=-(\frac{-6}{2(1)})=3$
$k=-(\frac{12}{2(1)})=-6$
The circle center is $(3,-6)$
The negative reciprocal of a slope of 2 os $-\frac{1}{2}$ since $2\times -\frac{1}{2}=-1$
The equation of the perpendicular line is
$y+6=-\frac{1}{2}(x-3)\Rightarrow y=-\frac{1}{2}x=4.5$
Set
$-\frac{1}{2}x-4.5=2x+10\Rightarrow 2.5x=-14.5\Rightarrow x=-\frac{29}{5}$
Plug $-\frac{29}{5}intoy=2(-295)+10=-\frac{8}{5}$
Use the distance equation:
$d=\sqrt{(3-(-\frac{29}{5}){)}^2+(-6-(-\frac{8}{5}){)}^2}$
$=\sqrt{(-\frac{44}{5}{)}^2+(-\frac{22}{5}{)}^2}=\frac{22\sqrt{5}}{5}\approx 9.8387units$
The shortest distance is about 9.8387 units.