The Equation of a Tangent Line to a Circle at a Point

Question

How do you determine the equation of the tangent line to the circle $(x-4)^2 + (y+3)^2 = 25$ at $P(8,-6)$?

Thank you in advance.

Answer 1

hint: $2(x-4) + 2(y+3)y' = 0 \to y'|_{(8,-6)} = -\dfrac{x-4}{y+3}|_{(8,-6)}= \dfrac{4}{3}$. Can you continue using the point-slope equation ?

Answer 2

Hint (without any differential calculus):

(1) The slope of the line joining points $\;(x_1,y_1)\;,\;(x_2,y_2)\;,\;\;x_1\neq x_2\;$ , is $\;\frac{y_1-y_2}{x_1-x_2} \;$.

(2) Two (non-vertical) lines with slopes $\;m_1\,,\,\,m_2\;$ are perpendicular to each other if and only if $\;m_1m_2=-1\;$ .

(3) The tangent line to a point on a circle is perpendicular to the circle's radius at that point .

Answer 3

The vector that has origin at the center of the circle $O$ and ends at $P$ is orthogonal to $\vec{PM}$ where $M(x,y)$ is a point of the tangent. So we have $\vec{OP}\cdot\vec{PM}=0$ and this means $$(8-4)(x-8)+(-6+3)(y+6)=0$$ and this gives the tangent equation $$4x-3y=50$$

Answer 4

we will use the quadratic equation, in particular, the fact that $ax^2 + bx+c = 0$ has repreat roots if and only if the discriminant $b^2 - 4ac = 0.$

pick a line through $P=(8, -6)$ with slope $m$ given by $$y = m(x-8) - 6 \tag 1$$ this line cuts the circle at $$(x-4)^2 + \left(m(x-8) - 6 + 3 \right)^2 = 25$$ it is easier to make change a variable here $x - 8 = u.$ with that we get the quadratic equation $$0 = (u+4)^2 + (mu-3)^2 - 25 = (1 + m^2)u^2 +(8-6m)u$$ and the discriminant is $$(8 - 6m)^2 = 0 \to m = \frac 43$$

the tangent line is $$y = \frac 43 (x-8) + 6 $$

ps. now i can see that the point $(8, -6)$ is on the circle. had i noticed that it would have far easier to find the tangent to a circle at a point on it than from a point outside of it.

Answer 5

Without calculus, and simply noting that the equation of the circle suggests an elementary application of Pythagoras may be helpful, draw a diagram.

In particular note that the point $(0,0)$ lies on the circle and that the point $(8,-6)$ is diametrically opposite.

The tangent is perpendicular to the diameter, and you can either use this to get the slope, or use similar triangles to find the $x$-intercept of the tangent. In the first case you have a point and a gradient. In the second case you have two points. In either case you have enough to determine the equation of the tangent.