Circles
Tangent Condition
Grade None
Question:
<p>Let C be the circle of radius unity centred at the origin. If two positive numbers \(x_1\) and \(x_2\) are such that the line passing through \((x_1, -1)\) and \((x_2, 1)\) is tangent to C, then:</p>
<p>(a) \(x_1 x_2 = 1\)</p>
<p>(b) \(x_1 x_2 = -1\)</p>
<p>(c) \(x_1 + x_2 = 1\)</p>
<p>(d) \(4x_1 x_2 = 1\)</p>
Step-by-Step Solution
Key Concept: A line is tangent to a circle if and only if the perpendicular distance from the center to the line equals the radius. Use this distance formula with the constraint that the line passes through two given points.
<p><strong>Step 1: Set up the line equation</strong></p><p>The line passes through points $(x_1, -1)$ and $(x_2, 1)$. Using the two-point form:</p><p>$$\frac{y - (-1)}{x - x_1} = \frac{1 - (-1)}{x_2 - x_1} = \frac{2}{x_2 - x_1}$$</p><p>Simplifying: $(y + 1)(x_2 - x_1) = 2(x - x_1)$</p><p>Expanding: $(x_2 - x_1)y + (x_2 - x_1) = 2x - 2x_1$</p><p>Rearranging: $2x - (x_2 - x_1)y - (x_2 + x_1) = 0$</p></p><p><strong>Step 2: Apply tangency condition</strong></p><p>The distance from origin $(0,0)$ to the line $2x - (x_2 - x_1)y - (x_2 + x_1) = 0$ must equal 1:</p><p>$$\frac{|2(0) - (x_2 - x_1)(0) - (x_2 + x_1)|}{\sqrt{4 + (x_2 - x_1)^2}} = 1$$</p><p>$$\frac{x_2 + x_1}{\sqrt{4 + (x_2 - x_1)^2}} = 1$$</p><p>(Since $x_1, x_2 > 0$, we have $x_2 + x_1 > 0$)</p></p><p><strong>Step 3: Square both sides</strong></p><p>$$(x_2 + x_1)^2 = 4 + (x_2 - x_1)^2$$</p><p>$$x_2^2 + 2x_1x_2 + x_1^2 = 4 + x_2^2 - 2x_1x_2 + x_1^2$$</p><p>$$2x_1x_2 = 4 - 2x_1x_2$$</p><p>$$4x_1x_2 = 4$$</p><p>$$x_1x_2 = 1$$</p><p><strong>∴ Answer: A</strong></p>
Correct Answer: A