Math Help - Find point on circle such that angle is minimum?

1. Find point on circle such that angle is minimum?

Hello everyone!

Find a point P on the circle $x^2 + y^2 - 4x - 6y + 9 = 0$ such that $\angle\mbox{POX}$ is minimum, where $\mbox{O}$ is the origin and $\mbox{OX}$ is the x-axis.

I tried sketching the curve but that didn't help. Parametric coordinates are the key? Or any other approach?

2. Originally Posted by fardeen_gen
Hello everyone!

Find a point P on the circle $x^2 + y^2 - 4x - 6y + 9 = 0$ such that $\angle\mbox{POX}$ is minimum, where $\mbox{O}$ is the origin and $\mbox{OX}$ is the x-axis.

I tried sketching the curve but that didn't help. Parametric coordinates are the key? Or any other approach?
Hi fardeen_gen,

you can locate the circle by finding it's centre and radius.
Then it's possible to work geometrically from that.

$x^2+y^2-4x-6y+9=0$

complete the squares for x and y

$x^2-4x\color{blue}+4\color{black}+y^2-6y+9\color{blue}-4\color{black}=0$

$(x-2)^2+(y-3)^2=4=r^2$

hence the radius is 2, centre is (2,3)

Referring to the attachment, there are identical triangles
with perpendicular sides of lengths 2 and 3.

Hence

$tan\theta=\frac{2}{3}$

$\theta=tan^{-1}\left(\frac{2}{3}\right)=33.69^o$

$2\theta=67.38^o$

Hence the angle at the origin in the triangle on the x-axis is

$90-67.38=22.62^o$

We can now find the x and y co-ordinates of P.

$cos22.62^o=\frac{x}{3}\ \Rightarrow\ x=3cos22.62^o=2.77$

$sin22.62^o=\frac{y}{3}\ \Rightarrow\ y=3sin22.62^0=1.154$

3. I hadn't thought about making a line passing through the centre. Very nice

Other users are welcome to solve it by alternate methods.

4. Here's another way,

if we differentiate the circle equation,
we will have a function describing the tangent slope
at all points (x,y) on the circle.

$x^2+y^2-4x-6y+9=0$

$\frac{d}{dx}\left(x^2+y^2-4x-6y+9\right)=0$

$2x+2y\frac{dy}{dx}-4-6\frac{dy}{dx}=0$

$\frac{dy}{dx}(2y-6)=4-2x$

$\frac{dy}{dx}=\frac{4-2x}{2y-6}=\frac{2-x}{y-3}$

this is the tangent slope at any point (x,y) on the circle circumference.

We are looking for the line of the form y=mx that passes through
the origin, since y=mx+c has c=0.

This is the tangent that goes through the origin.

Hence

$y=\left(\frac{x-2}{3-y}\right)x$

If we rearrange this we can place the x and y into the circle equation.

$y=\frac{x^2-2x}{3-y}\ \Rightarrow\ 3y-y^2=x^2-2x$

$2x+3y=x^2+y^2$

hence we can replace the $x^2+y^2$ in the circle equation

obtaining a linear equation in x and y at the point of tangency.

Therefore

$2x+3y-4x-6y+9=0$

$-2x-3y+9=0$

$3y=-2x+9$

$y=-\frac{2}{3}x+3$

This line goes through the point of tangency but crosses the y-axis at y=3.
This is because there are 2 tangents to the circle that pass through the origin..
the second being the y-axis.
Hence our resulting linear equation is the line passing through both points of tangency
for lines that go through the origin.

We want the line that goes through this point and the origin.

Hence

$y=\frac{x^2-2x}{3-y}\ \Rightarrow\ -\frac{2}{3}x+3=\frac{x^2-2x}{3+\frac{2}{3}x-3}$

$=\frac{x^2-2x}{\frac{2}{3}x}$

$\left(-\frac{2}{3}x+3\right)\frac{2}{3}x=x^2-2x$

$\frac{-4}{9}x^2-\frac{9}{9}x^2=-4x$

$-\frac{13}{9}x^2=-4x$

$x^2=\frac{36}{13}x=2.77x$

$x(x-2.77)=0$

The x co-ordinate is 2.77

Hence we can find the y co-ordinate and write the tangent slope and equation.