Question

If the line $y = mx + 2$ touches the circle ${x^2} + {y^2} = 1$, Then the positive value of $m$ is
(1) $\sqrt 3 $
(2) $\dfrac{1}{{\sqrt 3 }}$
(3) 1
(4) $\sqrt 2 $

Answer 1

Hint: First of all, find the coordinates of center and the radius of the given circle. Then, we know that the tangent is always perpendicular to the radius at the point of the contact. Hence, use the distance formula from a point to a line to form an equation in $m$. Solve the equation to get the required answer.

Complete step by step answer:

We are given that the line $y = mx + 2$ touches the circle ${x^2} + {y^2} = 1$ , then we can say that the line is tangent to the circle.
We also know that the line from the centre is perpendicular to the tangent.
First of all, we will find the coordinates of the centre of the circle and the radius of the circle.
Compare the given equation of the circle with the general equation of the circle which is ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$, where $\left( {h,k} \right)$ are the coordinates of the centre of the circle and $r$ is the radius of the circle.
Hence, the coordinates of the centre of the circle are $\left( {0,0} \right)$ and the radius of the circle is 1 unit.
Since the line is tangent to the circle, the perpendicular distance from the centre is equal to 1.
We know that the perpendicular distance from the centre is given as $D = \dfrac{{\left| {a{x_1} + b{y_1} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}$, where $\left( {{x_1},{y_1}} \right)$ are the coordinates of the point and $ax + by + c = 0$ is the equation of the line.
We are given that the equation of the line is $y = mx + 2$, which can be written in standard form as $ - mx + y - 2 = 0$
Hence, on substituting the values in the above formula, we get, $1 = \dfrac{{\left| {\left( { - m} \right)\left( 0 \right) + 1\left( 0 \right) - 2} \right|}}{{\sqrt {{m^2} + {1^2}} }}$
We will now solve the above equation using the property $\left| a \right| = a$ and $\left| { - a} \right| = a$
$
  1 = \dfrac{{\left| { - 2} \right|}}{{\sqrt {{m^2} + 1} }} \\
   \Rightarrow 1 = \dfrac{2}{{\sqrt {{m^2} + 1} }} \\
$
Cross multiply to solve for the value of $m$
$\sqrt {{m^2} + 1} = 2$
Squaring both sides,
$
  {m^2} + 1 = 4 \\
   \Rightarrow {m^2} = 3 \\
$
Taking square-root on both sides, $m = \sqrt 3 $

Hence, option A is correct.

Note: Here, the perpendicular distance will be equal to the radius because the line is tangent to the circle. Similarly, the perpendicular distance will be less than radius if the line cuts the circle at two points. Also, the $m$ represents slope when the equation of the line is of the form $y = mx + c$.