In an equation,if any variable is not equal to zero,we can multiply or divide the equation with it,does this also imply that it is also undefined?

Question

Find the equation of tangents through the origin to the circle $x^2+y^2-2rx-2hy+h^2=0.$

Let the given tangent be $y=mx$

The perpendicular distance of this tangent from the centre of circle must be equal to $r$.

Now I know that one tangent will surely be $x=0$.

And while solving the equation I mentioned,there’s a term of $m^2r^2$ on both sides of the equation.

Since m is not equal to zero,I can subtract these two equal terms. But that means that one value of m can be zero,but that would mean the line is $y=0$ which is wrong unless $r=h$

How do I solve this enigma?

Answer 1

Not all straight lines through the origin may be given by the equation $y=mx$, you’ve mentioned the straight line given by $x=0$. For that particular line no $m$ exists at all.

For the tangent given by $x=0$ you will find $y=h$ and the distance from $(0,h)$ and the center $(r,h)$ is indeed $r$.

Answer 2

You have found that one tangent is the y-axis.

To find the other one substitute y=mx in your equation $$x^2+y^2-2rx-2hy+h^2=0 $$ to get $$ (1+m^2 )x^2 -2(r+hm)x+h^2 = ax^2+bx+c =0 $$

For the tangency you need this equation to have a double root.

Thus you solve for $$b^2-4ac =0$$

and find $$m= \frac {h^2-r^2}{2rh}$$

Answer 3

I see two completely different ways to approach this equation. One is to complete the square for $x$ and also for $y$: $$ (x - r)^2 + (y - h)^2 = x^2 + y^2 - 2rx - 2hy + h^2 + r^2, $$ and therefore the equation of the circle can equivalently be written $$ (x - r)^2 + (y - h)^2 = r^2. $$ Now you just need to find the two lines through the origin that are at distance $r$ from the point $(r,h).$

Another way is to write the equation of a line through the origin and plug this into the equation of the circle. A line that intersects the circle twice will produce an equation with two solutions; a line that does not intersect the circle will produce an equation with no solutions; a tangent line will produce an equation with exactly one solution.

A line through the origin can be given by either $y = mx$ or by $x = 0.$ If you use these equations you have to treat them as two separate cases.

For the case $x = 0,$ supposing you did not already see that it is tangent to the circle, you can put $x = 0$ in the equation of the circle. The result is $$y^2-2hy+h^2=0.$$ This has exactly one root, $y = h,$ and therefore $x = 0$ is a tangent to the circle.

For the case $y = mx,$ you get a quadratic in $x$ and must set $m$ so that the discriminant is zero, as shown in another answer that was posted while I was writing this one.

Answer 4

It looks like you took the formula for the distance of a point from a line, squared it an rearranged to arrive at the equation $$(mr-h)^2 = (1+m^2)r^2$$ which expands into $$m^2r^2-2mrh+h^2=r^2+m^2r^2.$$ What seems to be troubling you is that the appearance of an $m^2$ term suggests that this equation might have two solutions, but that’s not the case. You could replace $m^2r^2$ on both sides with any expression whatsoever that involves $m$ and they would still cancel, leaving you with a linear equation in $m$. Adding the same thing to both sides of an equation doesn’t remove any solutions in the way that, say, dividing by some quantity or taking square roots can.

Another way to reassure yourself that you’re not omitting any solutions is this: there are at most two tangents through a circle through any point and you already know that $x=0$ is one of the tangents. However, that line can’t be expressed in the form $y=mx$ since its slope is infinite, so whatever equation you come up with to solve for $m$ is going to have at most one valid solution.

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Since I can’t resist joining the party, I’ll add another alternate solution here without detailed explanation. For any nondegenerate conic with matrix $C$, the tangents through a point $\mathbf p$ are given by the degenerate conic $\mathbf p_\times^TC^{\tiny\triangle}\mathbf p_\times$. Here $$C = \begin{bmatrix}1&0&-r\\0&1&-h\\-r&-h&h^2\end{bmatrix}, C^{\tiny\triangle} = \begin{bmatrix}0&hr&r\\hr&h^2-r^2&h\\r&h&1\end{bmatrix}, \mathbf p_\times = \begin{bmatrix}0&-1&0\\1&0&0\\0&0&0\end{bmatrix} $$ so $$\mathbf p_\times^TC^{\tiny\triangle}\mathbf p_\times = \begin{bmatrix}r^2-h^2&hr&0\\hr&0&0\\0&0&0\end{bmatrix},$$ which corresponds to the equation $$(r^2-h^2)x^2+2hrxy = 0.$$ This is easily factored into $x\,((r^2-h^2)x+2hry)=0$, the product of equations of the two tangent lines.