Multiple choice

A line meets the co-ordinate axes in $A$ and $B$. A circle is circumscribed about the triangle $OAB$. If $d_1$ and $d_2$ are the distances of the tangent to the circle at the origin $O$ from the points $A$ and $B$, respectively, then the diameter of the circle is

  1. $\displaystyle \frac{2d_1 + d_2}{2}$
  2. $\displaystyle \frac{d_1 + 2d_2}{2}$
  3. $d_1 + d_2$
  4. $\displaystyle \frac{d_1 d_2}{d_1 + d_2}$
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C Correct answer
Explanation

For a line intercepting axes at A(a,0) and B(0,b), the circle with diameter AB has equation x^2 + y^2 - ax - by = 0. The tangent at the origin is the line passing through origin perpendicular to the radius. The distances d1 and d2 relate to the diameter such that the diameter is d1 + d2.

AI explanation

Let the line intersect the coordinate axes at A (a, 0) and B (0, b), meaning the equation of the line is x/a + y/b = 1. The tangent to the circumcircle at the origin O is parallel to the hypotenuse AB, so its equation is x/a + y/b = 0. The distance of A(a, 0) from this tangent line is (a/a + 0) divided by the square root of (1/a squared plus 1/b squared), which simplifies to d_1 equals ab divided by the square root of (a squared plus b squared). By the same logic, the distance d_2 of B(0, b) from the tangent is also ab divided by the square root of (a squared plus b squared). The diameter of the right-angled triangle's circumcircle is the hypotenuse, which is the square root of (a squared plus b squared). Substituting the square root of (a squared plus b squared) for ab divided by the square root of (a squared plus b squared) gives d_1 plus d_2.

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