# 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 :)

• How about substitution? Replace your "y" term of the circle by the equation of the line. This gives you a quadratic equation. there are two solutions. One of them gives the x-coodinate of the point you are looking for. Nov 19, 2013 at 20:02
• @imranfat: Nice point. However I find it important to mention what you are doing, not just how to get to "the right number". Basically you have two equations of two unknowns and you want to find the set of points in R2 that satisfies both equations and one way to solve such a system of equations is substitution.
Nov 19, 2013 at 21:25

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.

• +1 I didn't even notice your post before, @amWhy...hehe. Nov 19, 2013 at 20:12
• @amWhy: Deserves another TU +1 Nov 20, 2013 at 1:32

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.)

You have the equation for the circle, simply insert $y=\frac{9}{7}x$ into it and solve for $x$.

Draw a picture: in our special case the distance is $\sqrt{7^2+9^2}-7$.

• In our course we don't really draw, so the answers I could use is substituting y with 9/7x like others said or your answer. In our book we have an example for a similar question, and the answer was similar to the one you mentioned (distance between origin and center minus h). But I don't understand how we got it, could you clarify it more please? (Thanks everyone :) Nov 19, 2013 at 21:17
• Not drawing pictures might be considered as a bad habit, hence just draw a picture of that specific circle: you'll see that that (provided the origin is not inside of the circle) you may calculate the distance between origin and center minus radius. Nov 20, 2013 at 11:23

Substitute in the circle's equation and get the $\;x$-coordinate, and then the $\;y$-coordinate:

$$y=\frac97x\implies (x-7)^2+\left(\frac97x-9\right)^2=49\iff$$

$$(x-7)^2+\frac{81}{49}(x-7)^2=49\iff \frac{130}{49}(x-7)^2=7^2\iff$$

$$(x-7)^2=\frac{7^4}{130}\iff x-7=\pm\frac{7^2}{\sqrt{130}}\iff\;\ldots$$