Question

If $\frac{x}{\alpha }+\frac{y}{\beta }=1$ touches the circles $x^2+y^2=a^2$, then point $\left(\frac{1}{\alpha },\frac{1}{\beta }\right)$ lies on

• A. Straight line
• B. Circle
• C. Parabola
• D. Ellipse

Answer 1

The correct option is B

Circle

Explanation for the correct option.

$\frac{x}{\alpha }+\frac{y}{\beta }=1$ touches the circle $x^2+y^2=a^2$, it means it is a tangent.

So, its distance from center $\left(0,0\right)$ will be equal to the radius, that is $a$.

$$\begin{gathered} \Rightarrow \left|\frac{{\displaystyle \frac{0}{\alpha }}+{\displaystyle \frac{0}{\beta }}-1}{\sqrt{{\displaystyle \frac{1}{{\alpha }^2}}+{\displaystyle \frac{1}{{\beta }^2}}}}\right|=a \\ \Rightarrow \left|\frac{{\displaystyle -1}}{\sqrt{{\displaystyle \frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}}}}\right|=a \\ \Rightarrow \frac{1}{\sqrt{\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}}}=a \end{gathered}$$

By squaring both sides, we get

$\begin{array}{rcl}\frac{1}{{\displaystyle \frac{1}{{\alpha }^2}}+{\displaystyle \frac{1}{{\beta }^2}}}& =& a^2\\ \Rightarrow \frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}& =& \frac{1}{{\displaystyle a^2}}\\ & \Rightarrow & {\left(\frac{1}{\alpha }\right)}^2+{\left(\frac{1}{\beta }\right)}^2={\left(\frac{1}{a}\right)}^2\end{array}$

Therefore, the point $\left(\frac{1}{\alpha },\frac{1}{\beta }\right)$ lies on the circle, whose equation is $x^2+y^2={\left(\frac{1}{a}\right)}^2$.

Hence, option B is correct.