Multiple choice

A circle of radius r is concentric with $\dfrac{x^2}{a^2}$ + $\dfrac{y^2}{b^2}$ = 1. The slope of a common tangent to them is

  1. $\sqrt{r^2 b^2}$
  2. $\sqrt{r^2 a^2}$
  3. $\sqrt{r^2 + b^2}$
  4. $\sqrt{\dfrac{b^2-  r^2}{r^2 -a^2}}$
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D Correct answer
Explanation

The condition for a line y = mx + c to be tangent to the ellipse x^2/a^2 + y^2/b^2 = 1 is c^2 = a^2m^2 + b^2. For a circle x^2 + y^2 = r^2, the condition is c^2 = r^2(1 + m^2). Equating the two: a^2m^2 + b^2 = r^2 + r^2m^2. Solving for m^2: m^2(a^2 - r^2) = r^2 - b^2, so m = sqrt((b^2 - r^2) / (r^2 - a^2)).

AI explanation

Assume the common tangent has the equation y = mx + c. For it to be tangent to the circle y^2 = r^2(1 + m^2), the condition c^2 = r^2(1 + m^2) must hold true. For the same line to be tangent to the ellipse x^2/a^2 + y^2/b^2 = 1, the condition c^2 = a^2*m^2 + b^2 must be satisfied. Equating both expressions for c^2 gives r^2 + r^2*m^2 = a^2*m^2 + b^2, which rearranges to m^2 = (b^2 - r^2) / (r^2 - a^2). Thus, the slope is the square root of (b^2 - r^2) / (r^2 - a^2).

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