# Thread: Equation of Circle Tangent

1. ## Equation of Circle Tangent

To begin with, I solved this problem, and got the correct answer. The reason I'm posting is because the solution manual uses a different method, and I can't follow what they're doing. Please explain their solution.

Problem:
For what values of k is the straight line $y = kx + 1$ a tangent to the circle with center $(5,1)$ and radius 3?

My solution:
The distance from the y-intercept (0,1) to the center of the circle is 5, and the radius is 3. Since the tangent is perpendicular to the radius, a 3-4-5 triangle is created. Dropping an altitude in this triangle gives us a change-in-y, change-in-x triangle for the tangent line. Since it is similar to the large triangle, it also has an opposite/adjacent ratio of $\frac{3}{4}$, which is the slope of our tangent line. Similarly, there will be a symmetric tangent line with a slope of $\frac{-3}{4}$.

Their solution:

$
\frac{|5k-1+1|}{\sqrt{k^2+1}} = 3
$

...(algebraic work that I can follow)...

$
k = \pm\frac{3}{4}
$

Where is their starting equation coming from? Is this some distance from a line to a point equation that I've never seen?

2. Originally Posted by Henderson
Problem:
For what values of k is the straight line $y = kx + 1$ a tangent to the circle with center $(5,1)$ and radius 3?
Their solution:
$
\frac{|5k-1+1|}{\sqrt{k^2+1}} = 3
$

...(algebraic work that I can follow)...
$
k = \pm\frac{3}{4}
$

Where is their starting equation coming from? Is this some distance from a line to a point equation that I've never seen?
The distance from the point $(p,q)$ to the line $Ax+By+C=0$ is given as:
$\dfrac{|Ap+Bq+C|}{\sqrt{A^2+B^2}}$

3. Thanks- that's definitely the equation they're using. Can you point me toward a derivation of that formula? I've never seen it before.

4. Originally Posted by Henderson
Thanks- that's definitely the equation they're using. Can you point me toward a derivation of that formula? I've never seen it before.
As you can see this best done using vectors.