Question

If the line $y\cos \alpha =x\sin \alpha +a\cos \alpha$ is a tangent to the circle $x^2+y^2=a^2$, then

• A. ${\sin }^2\alpha =1$
• B. ${\cos }^2\alpha =1$
• C. ${\sin }^2\alpha =a^2$
• D. ${\cos }^2\alpha =a^2$

Answer 1

The correct option is B

${\cos }^2\alpha =1$

Explanation for the correct option:

Compute the perpendicular distance between the centre of the given circle and the line

In the question, an equation of the circle $x^2+y^2=a^2$ and an equation of the line $y\cos \alpha =x\sin \alpha +a\cos \alpha$ is given.

Rewrite the given equation of line as follows: $x\sin \alpha -y\cos \alpha +a\cos \alpha =0$

• The standard equation of a circle is given by: ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$, where $(x,y)$ is the general point on the circle, $(h,k)$ are the coordinate of the centre and $r$ is the radius of the circle.
• The perpendicular distance $d$ between a line $Ax+By+C=0$ and a point $(x_1,y_1)$ is given by $d=\left|\frac{A{x}_1+B{y}_1+C}{\sqrt{A^2+B^2}}\right|$.
• The perpendicular distance between a circle and its tangent is equal to the radius of the circle.

So, the centre of the given circle is at $(0,0)$ and the radius is $a$.

Therefore, the equation for the perpendicular distance between the given circle and the given line is as follows:

$$\begin{gathered} \left|\frac{\left(0\right)\sin \alpha -\left(0\right)\cos \alpha +a\cos \alpha }{\sqrt{{\sin }^2\alpha +{\left(-\cos \alpha \right)}^2}}\right|=a \\ \Rightarrow \left|\frac{a\cos \alpha }{\sqrt{{\sin }^2\alpha +{\cos }^2\alpha }}\right|=a \\ \Rightarrow \left|a\cos \alpha \right|=a \\ \Rightarrow a^2{\cos }^2\alpha =a^2 \\ \Rightarrow {\cos }^2\alpha =1 \end{gathered}$$

Therefore, If the line $y\cos \alpha =x\sin \alpha +a\cos \alpha$ is a tangent to the circle $x^2+y^2=a^2$, then ${\cos }^2\alpha =1$.

Hence, option (B) is the correct answer.