Question

For any two nonzero real numbers \(a\) and \(b\), if the line
\[ \frac{x}{a}+\frac{y}{b}=1 \] is a tangent to the circle
\[ x^2+y^2=1, \] then which of the following is true?

• A. \(\left(\dfrac1a,\dfrac1b\right)\) lies inside the circle
• B. \((a,b)\) lies inside the circle
• C. \(\left(\dfrac1a,\dfrac1b\right)\) lies on the circle
• D. \((a,b)\) lies on the circle

Hint:

For a line to be tangent to a circle, the perpendicular distance from the center to the line must equal the radius.

Answer 1

The Correct Option is C

Step 1: Put the line in general form.
Multiply $\dfrac{x}{a}+\dfrac{y}{b}=1$ by $ab$ to get $bx+ay-ab=0$.
Step 2: Recall the tangent condition.
The circle $x^2+y^2=1$ has centre $(0,0)$ and radius $1$. Tangency means the distance from the centre equals $1$.
Step 3: Apply the distance formula.
$\dfrac{|ab|}{\sqrt{a^2+b^2}}=1$.
Step 4: Square and simplify.
$a^2b^2=a^2+b^2$.
Step 5: Divide by $a^2b^2$.
$1=\dfrac{1}{b^2}+\dfrac{1}{a^2}$, that is $\dfrac{1}{a^2}+\dfrac{1}{b^2}=1$.
Step 6: Interpret the result.
The point $\left(\dfrac1a,\dfrac1b\right)$ satisfies $x^2+y^2=1$, so it lies on the circle.
\[ \boxed{\left(\dfrac1a,\dfrac1b\right)\ \text{lies on the circle}} \]