MHB How Can I Find the Equation of the Dotted Tangent Line of a Circle?

[Yankel]

Dear all,

Attached is a picture of a circle.

View attachment 9133

The lower tangent line is y=0.5x. The center of the circle is M(4,7) while the point A is (3,6).

I found the equation of the circle, it is:

$(x-4)^{2}+(y-7)^{2}=20$

and I wish to find the dotted tangent line. I know that it is parallel to the lower one, therefore, the slope is the same. I can't find the tangent point.

Can you give me a hint please ?

 

[LCKurtz]

Dear all,

Attached is a picture of a circle.
The lower tangent line is y=0.5x. The center of the circle is M(4,7) while the point A is (3,6).

I found the equation of the circle, it is:

$(x-4)^{2}+(y-7)^{2}=20$

and I wish to find the dotted tangent line. I know that it is parallel to the lower one, therefore, the slope is the same. I can't find the tangent point.

Can you give me a hint please ?

You know $A=(3,6)$, and a direction vector along the lower tangent line is $\langle 2,1\rangle$. Can you find a vector perpendicular to that vector and pointed upwards towards the other tangent? And make it $\sqrt{20}$ units long? Then just add its components to $(3,6)$ to get the other tangent point.

[Edit] I notice that your point A doesn't satisfy the equation of your circle, so check your work so far.

 

[Yankel]

I am sorry, I did not mention this. The problem is in analytical geometry, no vectors are allowed

 

[I like Serena]

I am sorry, I did not mention this. The problem is in analytical geometry, no vectors are allowed

Since the tangent line is parallel, it must be of the form $y=\frac 12 x + b$ for some constant $b$, which is also the intersection with the y-axis.

Substitute in the circle equation, which becomes a quadratic equation.
Since it is a tangent it has exactly 1 point on the circle.
This happens when the square root part of the quadratic solution is zero.
From there we can find $b$.

 

[Opalg]

Something is wrong here. If a circle has centre at $(4,7)$ and goes through the point $A$ at $(3,6)$, then the tangent at $A$ will not go through the origin. It looks to me as though $A$ should be the point $(6,3)$ rather than $(3,6)$.

 

[Olinguito]

Hi Yankel.

As Opalg pointed out, there is a typo with the co-ordinates of the point A; it should be $(6,3)$ rather than $(3,6)$.

Let B with co-ordinates $(u,v)$ be the point opposite A on the circle. Then the line segment BA is perpendicular to the tangent line $y=\frac12x$ and so has gradient $-2$, i.e.
$$\frac{v-3}{u-6}\ =\ -2$$
from which we get
$$v\ =\ -2u+15\quad\ldots\boxed1.$$
Also $(u,v)$ lies on the circle, so
$$(u-4)^2+(v-7)^2\ =\ 20\quad\ldots\boxed2.$$
Substitute $v$ from $\boxed1$ into $\boxed2$ will give you a quadratic equation in $u$, which you can easily solve (knowing that $u=6$ is one solution).