# Straight line is tangent to the curve.

The straight line $y=mx+1$ is tangent to the curve $x^2+y^2-2x+4y=0$. Find the possible values of $m$.

My attempt

Substitute the $y=mx+1$ into the equation $x^2+y^2-2x+4y=0$. $$x^2+(mx+1)^2-2x+4(mx+1)=0$$ $$x^2+m^2x^2+2mx+1-2x+4mx+4=0$$ $$(1+m^2)x^2+6mx-2x+5=0$$ $$(1+m^2)x^2+(6m-2)x+5=0$$

I think what I did is wrong as I don't know how to continue from my steps. Can anyone explains it? Thanks

• If line $y=mx+1$ is a tangent to $F(x, y)=0$, where $F(x, y)$ is a polynom of degree 2, then $F(x, mx+1)=0$ have exactly one solution. Hence, discriminant is zero: $(6m-2)^2=4\cdot(1+m^2)\cdot5$. – Michael Galuza Jul 7 '15 at 16:24
• your method and computations are correct. Now what does thelast equation represent ? and what is the definition of a tangent to a curve?... – lmsteffan Jul 7 '15 at 16:24
• What type of curve is it? hint: add $1+4$ to both sides of the equation for the circle...oops! – John Joy Jul 7 '15 at 22:35
• @HarshKumar In case you were not aware, there is some discussion about whether the tag straight-lines is useful. As the creator of the tag, perhaps you would like to weigh in here. – pjs36 Jan 1 '17 at 19:17

If line $y=mx+1$ is a tangent to $F(x, y)=0$, where $F(x, y)$ is a polynom of degree $2$, then $F(x, mx+1)=0$ have exactly one solution. Hence, discriminant is zero: $$(6m-2)^2=4\cdot(1+m^2)\cdot5$$

• It is easier to observe that $z^2+y^2-2x+4y=0$ is the circle centered at $(1,-2)$ with radius $\sqrt{5}$. – wdacda Jul 7 '15 at 16:32
• @wdacda, easier for what? For this problem I can find $m$ immediately for any quadratic polynom. Geometry is cool, but in this case I prefer pure algebra. – Michael Galuza Jul 7 '15 at 16:36
• It seems to me that your argument is incomplete without knowing that $F(x,y)=0$ is a circle, or at least a complete argument would be more complicated for a general quadratic polynomial. In this setting I think that it is preferable to offer the simplest complete argument. This is just a personal preference. It is your solution, I did not mean to diminish it in any way. Sorry, if it appeared that way. – wdacda Jul 8 '15 at 4:16
• @wdacda, if $\deg F = 2$, line can intersect curve in 2 points (secant), 1 (tangent) or 0. In my university course of analytic geometry this facts were used without any arguments because they are trivial. – Michael Galuza Jul 8 '15 at 4:33
• What I see from your last comment is that your concept of trivial differs from mine. I do not think that a concept of a tangent line is trivial. So, anything that involves this concepts cannot be trivial, in my opinion. You did not formulate the fact from your course in analytic geometry as a formal mathematical statement, so I can only guess what you mean. I do not see how the polynomial $F(x,y) = y^2 - x$ and the line $y = 0$ fit into the trichotomy that you presented. – wdacda Jul 8 '15 at 16:56

Hint: Differentiating implicitly yields $$2x + 2y \frac{\mathrm{d}y}{\mathrm{d}x} - 2 + 4\frac{\mathrm{d}y}{\mathrm{d}x} = 0 \iff \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1-x}{y+2}$$

What can you say about gradient of tangent lines and derivatives of functions?

The general equation of a tangent to a circle $X^2 + Y^2 = a^2$ is given by $Y = mX \pm a\sqrt{1 + m^2}$.

For the purpose of your question, $X = x - 1$, $Y = y + 2$, $a = \sqrt{5}$. Also,

$$y = mx + 1 \\ \implies y + 2 = m(x - 1) + (m + 3) \\ \implies Y = mX + (m + 3).$$

Hence,

$$(m + 3) = \sqrt{5} \sqrt{1 + m^2}.$$

The rest is simple.