Finding the shortest distance between a point and a circle The question is "Find the shortest distance from the origin of the graph of the circle $x^2-14x+y^2-18y+81=0$ ".
I found the circle in the following form: $(x-7)^2+(y-9)^2=7^2$
Then I found the line that connects the origin $(0,0)$ and the center $(7,9)$, and it was $y=(9/7)x$
Now I want to find a point that is both on the circle AND on the line mentioned above, then find the distance between that point and $(0,0)$, but I don't know how to find that point.
My knowledge level is Preparatory mathematics. Thanks :)
 A: Solve the equation: $$(x-7)^2+\left[\underbrace{\left(\frac 97x\right)}_{y = \frac 97x}-9\right]^2 = 7^2$$ to find where the circle intersects your line (two points, one of which is closest to the origin). Call each point $(x_0, y_0)$.
To find the distance from the origin to each point, we know: $$d = \sqrt{(x^0 - 0)^2 + (y_0 - 0)^2} = \sqrt{x_0^2 + y_0^2}$$
Of the two points of intersection, choose the one for which $d$ is smallest.
A: If $y = \dfrac97 x$, and $(x-7)^2+(y-9)^2 = 7^2$, then $(x-7)^2+\left(\dfrac97 x-9\right)^2 = 7^2$.
That's a quadratic equation.  When you've found $x$, you can then multiply it by $\dfrac97$ to get $y$.
(Of course you get two solutions, since the line intersects the circle twice.)
A: You have the equation for the circle, simply insert $y=\frac{9}{7}x$ into it and solve for $x$.
A: Draw a picture: in our special case the distance is $\sqrt{7^2+9^2}-7$. 
A: Substitute in the circle's equation and get the $\;x$-coordinate, and then the $\;y$-coordinate:
$$y=\frac97x\implies (x-7)^2+\left(\frac97x-9\right)^2=49\iff$$
$$(x-7)^2+\frac{81}{49}(x-7)^2=49\iff \frac{130}{49}(x-7)^2=7^2\iff$$
$$(x-7)^2=\frac{7^4}{130}\iff x-7=\pm\frac{7^2}{\sqrt{130}}\iff\;\ldots$$
