How to determine if a line is tangent to a circle?

Question

In my book the question is:

Prove that line: $x - 3y = 5$ is a tangent to the circle: $x^2 + y^2 - 6x + 8y + 15 = 0$

P.S: If there is any general proof or derivation of it please let me know.I am also infant in mathematics.Only has learnt Line, Trigonometry of 9th and 10th grade and learning circles write now.

Answer 1

It has to touch the circle at one point, so plugging in x = 3y+5 in the second equation must have one solution for y.

You can also prove it by finding the line that passes through the point of intersection and the center of the circle, and then you can show that these two lines are prependicular to each other from the slopes.

Answer 2

Using implicit differentiation, we can find $\dfrac{dy}{dx}$ for the circle. Find the point of intersection for the line and the circle, and show that the slope of the tangent at that point is equal to the slope of the line.

\begin{align} x^2+y^2-6x+8y+15 &=0 \\ \dfrac{d}{dx}(x^2+y^2-6x+8y+15) &=\dfrac{d}{dx}\ 0 \\ 2x+2y\dfrac{dy}{dx}-6+8\dfrac{dy}{dx}& =0 \\ (2y+8)\dfrac{dy}{dx}&= 6-2x \\ \dfrac{dy}{dx}&=\dfrac{6-2x}{2y+8}=\dfrac{3-x}{y+4} \end{align}

Plug in $x=3y+5$ into the circle.

\begin{align} x^2+y^2-6x+8y+15 &=0 \\ (3y+5)^2+y^2-6(3y+5)+8y+15&=0 \\ 9y^2+30y+25+y^2-18y-30+8y+15&=0 \\ 10y^2+20y+10&=0 \\ y&=-1 \end{align}

Plug $y=-1$ into the line to get the point. \begin{align} x-3y &=5 \\ x-3(-1)&=5\\ x&=2 \Rightarrow \textrm{point of intersection } (2,-1) \\ \end{align}

Slope of the tangent of the circle at $(2,-1)$: \begin{align} \dfrac{dy}{dx}\bigg|_{(2,-1)}&= \dfrac{3-(2)}{(-1)+4} =\dfrac{1}{3}\\ \end{align}

Slope of the given line: \begin{align} x-3y&=5\\ -3y&=-x+5\\ y&=\dfrac{1}{3}x+5\\ m&=\dfrac{1}{3}\\ \end{align}

Thus ends the proof.

Answer 3

Every circle is tangent to a line if, by isolating a variable from the line function, and substituting the result you obtained with the same variable you obtained IN the line equation , you obtain a quadratic equation that has a null discriminant. Or even better, you can compute the distance of the center of the circle from the line and you'll have to find that this distance has the same value of the radius.

Regarding the latter, you have to verify that the center $(3,-4)$ is far from the line $\sqrt{10}$ (sorry, I have still to learn how to write in math format) by using the formula $d=\frac{\abs{ax _C+by_C+c}}{\sqrt{a^2+b^2}}$ with $ax+by+c=0$ as the equation of the line. The latter formula can used to compute a distance of a point $C$ from the line $ax+by+c=0$ in general. Applying in your case, we get $$\frac{10}{\sqrt{10}}$$ which is $\sqrt{10}$ as expected.

Answer 4

The easiest way is to examine the determinate of their joint quadratic equation.

Eliminate one variable, say, $x$, in the two equations to obtain,

$$y^2+2y+1=0$$

whose determinate is $\Delta = b^2-4ac = 0$. That is only one solution, meaning the two curves intersect at only one point and, therefore, tangential to each other.