Should we consider the line passing through the origin is having equal intercepts? I am a grade 11 student and I had the following question on my math test.
$x^2 +y^2+4x-8y+2=0$
is a circle and we have to find out the equations of the tangents of the circle that cuts off equal intercepts of the same sign from the axes of coordinates.
Naturally, I started with the equation being $\frac{x}{a}+\frac{y}{a}=1$ and found 2 equations. I then thought that I was assuming that the intercepts can never be 0 if I only used the formula.
So I then used the formula $y=mx$ to find two more equations and then kept the 4 equations as the answer. I didn't get any marks; so I gave back the copy for a recheck as I still , to some extent, believe that I was right.
Was I right about this? Or I should have kept the 1st two equations only??
Some people wanted to know the process of how I found the equation more descriptively. Here it is:
Given, $x^2 +y^2+4x-8y+2=0$ is a circle.
$\therefore$ centre, $C \equiv (-2,4)$, Radius, $R =\sqrt{18}$
Now, let the equation be $\frac{x}{a}+\frac{y}{a}=1$  where $a \neq 0$
$\implies x+y-a=0 \cdots (1)$.
$\therefore $ perpendicular distance to (1) from the centre
$ = \frac{|-2+4-a|}{\sqrt{1^2+1^2}} $ which should be equal to the radius if the line I tangent.
$\therefore \frac{|-2+4-a|}{\sqrt{1^2+1^2}} = \sqrt{18}\cdots (2)$
Solving 2 we get, $a=8,-4$
Again let,
Equation be $y-mx=0$ if the intercept is 0
Perpendicular distance from centre = $\frac{|2m+4|}{\sqrt{m^2+1}}=\sqrt{18}$[Radius] $\cdots (3)$
solving 3 we get, $m=1,-\frac{1}{7}$
Thus the required equations of tangent are:
$x+y-8=0$
$x+y+4=0$
$y=x$
$7y=x$

This can be solved in yet another  way
$(x+2)^2+(y-4)^2=18$
Transform $x+2\to X$ and $y-4\to Y$ then the equation of the circle becomes $X^2+Y^2=18$ , now condition for $Y=mX+c$ to be tangent is $c^2=r^2(m^2+1)$, where r = radius of the circle . Now they move cut equal intercept on $x$ and $y$ axes(Also $X$ and $Y$ axes), therefore $m=1$(Disregarding the case of $m=-1$ because of same sign condition), this gives 2 quadratics and hence the 4 equations in $X$ and$Y$ co-ordinates , now convert them back to normal coordinate and this gives you the answer ...
