How can I find the Equation of tangent line to a circle(x^2 + y^2 = 45^2) at an angle 60 degree with horizontal axis?

Question

How can I find the Equation of tangent line to a circle(x^2 + y^2 = 45^2) at an angle 60 degree with horizontal axis?

I tried finding the equation of line which comes out to be y = √3 x + c, and the equation of the circle x^2 + y^2 = 45^2.

But here I don't have c so it could be any line that is 60 degree with the horizontal. But I know for sure it should be a tangent line.

Answer 1

$m = \tan (60°)=√3$ is already known, where $m$ is the slope of the tangent line.

Circle: $x^2+y^2 =r^2$, $r =45.$

Differentiate with respect to $x:$

$2y\dfrac{dy}{dx} = -2x$; or

$m=√3= \dfrac{dy}{dx}= -x/y$.

Hence: $x = -√3y$ at point of tangency.

Now:

$x^2+y^2=r^2$ combining with

$x=-√3y$:

$3y^2 +y^2 = r^2$, or

$y= ^{+}_{-}r/2$,and

$x= ^{-}_{+}√3(r/2)$

Gives 2 tangent lines :

1) Line 1 passes through $x_1= -√3(r/2);$ $ y_1 = r/2$.

2) Line 2 passes through $x_2= √3(r/2), y_2=-r/2$

Can you find the respective $c$?.

Answer 2

The distance from the centre $(0,0)$ of the circle to the tangent is $\displaystyle \left|\frac{\sqrt{3}(0)-(0)+c}{\sqrt{(\sqrt{3})^2+(-1)^2}}\right|$. So,

\begin{align*} \frac{c}{2}=\pm 45 \end{align*}

Alternatively, you can substitute the equation of the tangent to the equation of the circle to obtain a quadratic equation (in $x$, or in $y$). The quadratic equation has a double root and its discriminant is $0$.