I was solving one question coordinate geometry when i encountered this. I had to find slopes of tangents from a point to a circle. I applied condition of tangency that any line y=mx + c is tangent to a circle
$x^2 + y^2= a^2$ if $c^2=(a^2)(1+m^2)$
When I was solving for $x^2 + y^2= 9$ and slopes of tangents from a point $(3,5)$ were asked.
I applied this and $c=-3m+5 h (-3m+5)^2=9(1+m^)$
This quadratic reduces down to linear equation because coefficients of m^2 on both sides cancel each other as you can see
$$9m^2 + 25 -30m = 9 +9m^2$$
$m$ comes out to be $\frac{16}{30}$ i.e. $\frac{8}{15}$
As we can easily see that one of the tangent is vertical to circle and parallel to y-axis and m tends to infinity. Is this the reason why the quadratic was reduced to linear equation? Explain me this. If this is the case how should i go about finding slopes of tangents to the conics where i can't figure out easily the figure on Cartesian plane? What precautions should be taken?
When a quadratic is reduced to a linear one, this means one of the roots requires special attention. In the case when finding slopes, that slope is most likely a vertical line with $m = \tan 90^0$.
Graph sketching is always the starting point in tackling analytical problems. In your example, it is not that difficult to see that one of the tangents from $(3, 5)$ to the circle $x^2 + y^2= 9$ is the vertical line $x = 3$.
Of course, the restriction of requiring $m \ne \tan 90^0$ (otherwise, the whole discussion becomes meaningless) should have been stated in the very beginning in the derivation of the condition for tangency. I have viewed through several analytic geometry texts and, strange enough, none of them have that mentioned/included.