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: Rewrite the line in standard form.
Given line is
\[ \frac{x}{a}+\frac{y}{b}=1 \]
Multiplying throughout by \(ab\),
\[ bx+ay-ab=0 \]
This is the standard form of the line.

Step 2: Use the tangent condition for the circle.
The circle is
\[ x^2+y^2=1 \] whose center is \((0,0)\) and radius is \(1\).
For the line
\[ bx+ay-ab=0 \] to be tangent to the circle, the perpendicular distance from the center to the line must equal the radius.
Thus,
\[ \frac{|ab|}{\sqrt{a^2+b^2}}=1 \]
Squaring both sides,
\[ \frac{a^2b^2}{a^2+b^2}=1 \]
\[ a^2b^2=a^2+b^2 \]

Step 3: Divide by \(a^2b^2\).
Since \(a,b\neq0\), divide both sides by \(a^2b^2\):
\[ 1=\frac1{b^2}+\frac1{a^2} \]
Rearranging,
\[ \frac1{a^2}+\frac1{b^2}=1 \]

Step 4: Interpret geometrically.
The point
\[ \left(\frac1a,\frac1b\right) \] satisfies
\[ x^2+y^2=1 \]
Hence, the point lies on the circle.

Step 5: Final conclusion.
Therefore,
\[ \boxed{\left(\frac1a,\frac1b\right)\text{ lies on the circle}} \]