Finding tangent to a circle with only one coordinate given

Question

So I came across this MCQ:

Which one of the following is the equation of a tangent to the circle $ x^2 + y^2 = 9: $

A. $ x = -1$

B. $ x = 4$

C. $ y = -4$

D. $ y =3$

E. $x = 0$

I know how to find the tangent to a circle given a coordinate for a point. I just find the slope of the normal for that point after checking whether it lies on the circle or not and then get the slope of tangent from that and plug it into $$ (y - y_1) = m(x - x_1) $$

I can't figure out how to go about doing it with only one coordinate given though.

Answer 1

This sort of question tests exam technique more than anything. You need to be able to quickly exclude and zero-in on choices for MCQs.

Here, the first step is recognising that you're dealing with a circle, centred at the origin with radius $3$.

Second step is recognising that all the choices represent vertical or horizontal lines. These always have equations of the form $y = c$ or $x=c$ where $c$ is a constant. There are no slanted lines, which always involve both $y$ and $x$ in a non-trivial manner in the equation.

The only vertical tangents to the circle are $x = - 3$ and $x =3$. The only horizontal tangents are $y = - 3$ and $y=3$.

The choice becomes obvious very quickly.

Answer 2

For all other options other than D, Gives two values or doesn't gives real solutions at all indicating a secant and non-contacting line respectively. For e.g. $x=0 => y =+3,-3$

And $x=4 => y =$ complex

It's only for $y=3$ that x becomes zero and hence a single point on graph - a tangent!