Distance Between Line and Circle Consider the circle $C$ given by the equation 
$$(x+5)^2 + (y−10)^2 = 15$$
and the line $L$ given by the equation 
$$y = \frac{1}{2}x−3.$$


*

*Find the distance between $C$ and $L$.

*Find the equation of the line perpendicular to $L$ that goes through the center of the circle $C$.

 A: *

*$\mathcal{C}$ has for equation $(x+5)^2+(y−10)^2=15$ meaning $\mathcal{C}$ has a $\sqrt{15}$ radius and a centre point that we will name $A$ with the following coordinates $A(-5;10)$.



• PROOF 1
Consider $\left(O,\vec{i},\vec{j}\right)$
  an orthonormal axis.
  One equation of the circle $\mathcal{C}$ of centre $A(x_A; y_A)$ and radius $R$ is
  $(x − x_A)^2 +(y − y_A)^2 = R^2$.

The shortest distance between $A$ and $L$ is the distance between A and its orthonormal projection $A'$. The normal vector of $L$ (denoted $\vec{n}$) and $\vec{A'A}$ are codirectional meaning that the scalar product between these two vectors is the product of their magnitude where $\vec{n}(a;b)$ because $(L) : ax+by+c=0$ : $$\left|\vec{A'A}\cdot\vec{n}\right|=||\vec{A'A}||\times||\vec{n}||$$

• THEOREM 1
Consider the slope $(d) : ax+by+c$.
  The normal vector is characterized by $\vec{u}(a;b)$. 
• PROOF 2
The scalar product between two vectors is : $$\vec{v}\cdot\vec{w}=||\vec{v}||\times||\vec{w}||\times cos\left(\vec{v},\vec{w}\right)$$
  When two vectors are codirectional, the angle of rotation between these two vectors is $0rad$, where $cos(0)=1$. Therefore :
$$\vec{v}\cdot\vec{w}=||\vec{v}||\times||\vec{w}||\times cos\left(\vec{v},\vec{w}\right)=||\vec{v}||\times||\vec{w}||\times1=||\vec{v}||\times||\vec{w}||$$

Knowing the length of a vector is determined by $\sqrt{x^2+y^2}$, thus we have, for $\vec{n}$ : $||\vec{n}||=\sqrt{a^2+b^2}$. So we have $\left|\vec{A'A}\cdot\vec{n}\right|=||\vec{A'A}||\sqrt{a^2+b^2}$. We're looking for the distance $A'A$ :
$$||\vec{A'A}||=\frac{\left|\vec{A'A}\cdot\vec{n}\right|}{\sqrt{a^2+b^2}}$$
The scalar product $\left|\vec{A'A}\cdot\vec{n}\right|$ can also be calculated thanks to the coordinates of these two vectors. 

• THEOREM 2
The scalar product is also the sum of the products of the $x$ and the $y$ coordinates. If $\vec{u}(x_u;y_u)$ and $\vec{v}(x_v;y_v)$, therefore : $$\vec{u}\cdot\vec{v}=x_ux_v+y_uy_v$$

Consider $\vec{A'A}(x_A-x_{A'};y_A-y_{A'})\leadsto\vec{A'A}(-5-x_{A'};10-y_{A'})$ and $\vec{n}(a;b)$, we then have :
$$ \left|\vec{A'A}\cdot\vec{n}\right|=\left|a(-5-x_{A'})+b(10-y_{A'})\right|$$
$$ \left|\vec{A'A}\cdot\vec{n}\right|=\left|-5a-ax_{A'}+10b-by_{A'}\right|$$
The point $A'$, being the orthogonal projection of $A$ on $L$, belongs to $L$, then it respects the equation of a slope, then $ax_{A'}-by_{A'}=c$. So :
$$ \left|\vec{A'A}\cdot\vec{n}\right|=\left|-5a+10b+c \right|$$
Finally we then have : $$||\vec{A'A}||=\frac{\left|-5a+10b+c \right|}{\sqrt{a^2+b^2}}$$
Knowing that $(L) : y=\dfrac{1}{2}x-3\Leftrightarrow x-2y-6=0$ with $a=1$, $b=-2$, $c=-6$, replacing we have :
$$||\vec{A'A}||=\frac{\left|-5\times1+10\times(-2)-6 \right|}{\sqrt{\left(1\right)^2+(-2)^2}}$$
$$||\vec{A'A}||=\frac{31}{\sqrt{5}}=\dfrac{31\sqrt{5}}{5}$$
$$A'A\approx13.86$$
The closest point of the circle is then at $A'A-R=\dfrac{31\sqrt{5}}{5}-\sqrt{15}=\dfrac{-\sqrt{5}\left(5\sqrt{3}-31\right)}{5}\approx10.0$


*A line has an equation of the form $y=ax+c$. We consider $(A'A) : y'=a'x+c'$ and $(L) : y=\dfrac{1}{2}x-3$. Two lines are perpendicular if and only if $aa'=-1$.


$$\frac{1}{2}a'=-1$$
$$a'=-2$$
We then have $(A'A) : y=-2x+c'$. Knowing that $A(-5;10)$, then $10=-2\times(-5)+c'$ meaning that $c'=0$, so :
$$(A'A) : y=-2x$$ is the perpendicular line of (L) going threw the center of $\mathcal{C}$.
A: For the first part, you can use the general formula for the distance between a point and a straight line
$$d=\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}},$$
considering the point $(x_0,y_0)$ and the line with equation in the form $ax+by+c=0$. Take the point to be the center of the circle.
For the second part, two lines with equations $y=a_1x+b_1$ and $y=a_2x+b_2$ are perpendicular if $a_1 a_2=-1$, so you can find the slope. Since you also know a point in the line (the center of the circle), you're done.
A: Geometrically, the shortest distance from a circle to a straight line is the from the center of the circle perpendicular to the line.  Here, the center of the circle is $(-5, 10)$ and a line through that point, perpendicular to the line $y- 10= -2(x+ 5)$ or $y= -2x$.
Replacing $y$ by that in the equation of the circle, $$(x+5)^2+ (-2x- 10)^2= x^2+ 10x+ 25+ 4x^2+ 40x+ 100= 15$$ or, $$\implies 5x^2+ 50x+ 110= 0$$
This quadratic equation has two solutions.  One is the point on the circle closest to the line, the other is the point on the circle farthest from the line.
