# how to find the equation of a tangent line to a circle, given its slope and the eq. of the circle?

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?)

• Hint : if $A(x_A, y_A)$ is the point both on your circle and your line, then $\overrightarrow{\Omega A} \cdot \overrightarrow{AA'} = 0$, with $A'(x_A+1, y_A-\frac 34)$ and $\Omega$ being the center of your circle. Then use the fact that $A$ belongs to you circle. – krirkrirk Mar 4 '18 at 11:02
• – lab bhattacharjee Mar 4 '18 at 11:03

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.

• @KingTut It was a typo. – Rebellos Mar 4 '18 at 11:12
• ok ................... – King Tut Mar 4 '18 at 11:12
• Okay so I find -3x - 4y + 11 = 0 as the equation of the tangent line.. would that be right? – jdoe123 Mar 4 '18 at 11:20
• @jdoe123 You should find two equations. There's an absolute value there. – Rebellos Mar 4 '18 at 11:21
• Oh yeah!! I forgot about that. All good :) Thank you ! – jdoe123 Mar 4 '18 at 11:24

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?

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 $$6=-\frac34(5)+h\implies h=\frac{39}4$$

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 $$\frac{2-p_1}{p_2-2} = -\frac{3}{4}$$

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 $h = -11/4$.

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.