So I have a circle: $(x-2)^2 + (y-2)^2 = 25$ and I have a tangent line to this circle, with a slope of $m= -3/4$.
I have to find the equation of the tangent line, so I know the radius of the circle is $r = 5$ and I wrote the equation of the tangent line as:
$$y = -3/4x + h$$
So now I have to find $y, x$ and $h$, but I don't know if I can just replace $x$ and $y$ with the center points? Or do I have to find the point-line distance (and why?)
On
put the equation $$y=-\frac{3}{4}x+h$$ into the circle equation solve this equation for $x$ and set the discriminant equal to Zero and compute $h$ the equation is given by $$\frac{1}{16} \left(16 h^2-24 h x-64 h+25 x^2-16 x-272\right)=0$$ solving this for $x$ we obtain $$x=\frac{4}{25}\left(2+3h\pm\sqrt{429+112h-16h^2}\right)$$ nowsolve the equation $$429+112h-16h^2=0$$ for $h$ Can you finish?
On
Calculus approach:
We have $$(x-2)^2 + (y-2)^2 = 25\implies y=2+\sqrt{25-(x-2)^2}$$ so slope is $$\frac{dy}{dx}=-\frac12(25-(x-2)^2)^{-1/2}\cdot(-2(x-2))=\frac{x-2}{\sqrt{25-(x-2)^2}}=-\frac34$$ giving $$(x-2)^2=\frac9{16}(25-(x-2)^2)\implies (x-2)^2=9\implies x=-1,5$$ This gives the two pairs $(x,y)=(-1,-2), (5,6)$. Your equation is $y=-\dfrac34x+h$ and substituting the first pair gives $$-2=-\frac34(-1)+h\implies h=-\frac{11}4$$ and substituting the second pair gives $$6=-\frac34(5)+h\implies h=\frac{39}4$$
On
With a bit of differential geometry: The circle $\mathscr C$ is the set of points $p=(p_1,p_2)\in\mathbb R^2$ such that $F(p) = (p_1-2)^2+(p_2-2)^2-25 = 0$. The tangent space $T_p\mathscr C$ is the set of vectors that are orthogonal to the gradient $\nabla F(p)$, and it is spanned by $v := (\partial F/\partial p_2,-\partial F/\partial p_1) = (p_2-2,2-p_1)$.
You want your point $p$ to be such that the slope of the tangent is $-3/4$, so the coordinates of $v$ are such that \begin{equation} \frac{2-p_1}{p_2-2} = -\frac{3}{4} \end{equation}
Pulling this into the defining equation of the circle, we get $$(p_1-2)^2+(p_2-2)^2 = \frac{25}{9}(p_1-2)^2 = 25 $$ so $|p_1 - 2| = 3$.
There are two solutions for $p$ : $p = (5,6)$ and $p = (-1,-2)$.
Both solutions must lie on the tangent line.
For the first one, we get $6 = -(3/4)5 + h$ so $h= 39/4$
And for the second one, we get $-2 = -(3/4)(-1) +h $ and $h = -11/4$.
On
For my circle whose centre lies on the origin the equation of it's tangent in slope form is given by:
$y=mx + r\sqrt{1+m^{2}}$
Where $r$ is the radius of the circle and $m$ is the slope of the tangent.
On
Here is the derivation for the equation of the line with slope $m$ that is tangent to a circle of radius $r$ centered on the origin. (If the circle was not centered on the origin, we can simply shift this equation.)
Let the required tangent line be $y=mx+b$ and the point of tangency be $(x_0, y_0)$. We need to find $b$ in terms of $m$ and $r$.
The slope of the tangent line $m$ must equal to the slope of the circle at $(x_0, y_0)$. So let's find the equation for slope of the circle at $(x_0, y_0)$.
The equation of the circle is $$x^2 + y^2 = r^2.$$
Implicitly differentiating, we have $$2x + 2y \frac{dy}{dx} = 0.$$
Rearranging we have $$\frac{dy}{dx} = -\frac{x}{y}.$$
So we have $$m = -\frac{x_0}{y_0}.$$
We now have these three equations:
All we need to do is to find $b$ in terms of $m$ and $r$.
Solving the second equation for $x_0$ we have $x_0 = -m y_0$. Substituting this to the first equation we have
$$b = y_0 - m x_0 = y_0 - m (-m y_0) = y_0 (1+m)^2.$$
Substituting $x_0 = -m y_0$ to the third equation we have $$m^2 {y_0}^2 + {y_0}^2 = r^2.$$
Solving for ${y_0}^2$ we have $$y_0^2 = \frac{r^2}{m^2+1}.$$
Taking the square root we get $$y_0 = \pm \frac{r}{\sqrt{m^2+1}}.$$
Therefore
$$b = \pm \frac{r}{\sqrt{m^2+1}} \cdot (1+m^2) = \pm r \sqrt{m^2+1}.$$
Note that we have two values for $b$, so there are actually two tangent lines with the same slope.
Hence, the tangent lines with slope $m$ for a circle of radius $r$ centered on the origin are given by
$$y = mx \pm r \sqrt{m^2+1}.$$
Finally, if the center of the circle was instead $(h, k)$, the equation of the tangent lines are
$$y-k = m(x-h) \pm r \sqrt{m^2+1}.$$
In your case, $(h, k) = (2,2)$, $r=5$, and $m = -3/4$. So the tangent lines are $$y-2 = -\frac{3}{4}(x-2) \pm 5\sqrt{\left(-\frac{3}{4}\right)^2 + 1}.$$
Which can be simplified to
$$y = -\frac{3}{4} x + \frac{39}{4}$$ and $$y = -\frac{3}{4} x - \frac{11}{4}.$$
On
Let the tangent line of the circle $(x-x_0)^2+(y-y_0)^2=R^2$ has slope $m$ and let the point of tangency be $P(x_T,y_T).$ Then the slope of $OP$ is $\tan\theta=-\frac1m$ where $O(x_0,y_0)$ is the center of the circle. Therefore, we have two solutions for $P$: $$P(x_T,y_T)=(x_0\pm\cos\theta,y_0\mp\sin\theta)=(x_0\pm\frac{mR}{\sqrt{m^2+1}},y_0\mp\frac{R}{\sqrt{m^2+1}}).$$ And we have two such tangent lines given by: $$y=mx-mx_0\mp\frac{m^2R}{\sqrt{m^2+1}}+y_0\mp\frac{R}{\sqrt{m^2+1}}$$ or $$y-y_0=m(x-x_0)\pm R\sqrt{m^2+1}.$$ OP's question: $x_0=y_0=2$, $R=5$ and $m=-\frac{3}{4}$. So, the tangent lines are $$y-2=-\frac34(x-2)\pm5\sqrt{\frac{9}{16}+1}$$ $$y=-\frac34 x+\frac{14}{4}\pm\frac{25}{4}.$$
The center of the circle can be figured out by the given equation of the circle and is the point : $$C(2,2)$$
But then, if the given line is tangent to your circle, it means that the distance from the center of the circle should be exactly $5$. Manipulating the line equation you have derived, we can yield : $4y + 3x + k = 0$ and then by solving the distance formula, you can yield the exact equation (there will be 2 parallel and diametrically opposite equations thus two tangent lines) :
$$ \left|\frac{Ax_0 + By_0 + Γ}{\sqrt{A^2 +B^2}} \right| = d(P,ε) \Rightarrow \left|\frac{4\cdot 2 + 3 \cdot 2 + k}{\sqrt{4^2+3^2}}\right| = 5$$
$$\Leftrightarrow$$
$$|8 + 6 + k| = 25 \Leftrightarrow \dots$$
Another approach would be substituting the line equation for $x$ and $y$ into your circle's equation and then demanding the equation to have a unique solution, since a tangent line will only have one common point with a circle.
Note : This only works for the case of the circle, when a tangent line can never have $2$ common points. This is not the case for other curves though.