Question

A line meets the co-ordinate axes in A and B. A circle is circumscribed about the triangle OAB. If d₁ and d₂ are the distances of the tangent to the circle at the origin O from the points A and B respectively, the diameter of the circle is :

• A. $\frac{2d_1+d_2}{2}$
• B. $\frac{d_1+2d_2}{2}$
• C. $d_1 + d_2$
• D. $\frac{d_1d_2}{d_1+d_2}$

Answer 1

Answer: (C)

$d_1 + d_2$

Answer 2

Let the line meeting the co-ordinate axes at A and B be $\frac{x}{a} + \frac{y}{b} = 1$. Thus, the coordinates of A are $(a, 0)$ and the coordinates of B are $(0, b)$. O is the origin $(0, 0)$. The triangle OAB is a right-angled triangle with the right angle at O. The circle circumscribed about $\triangle OAB$ has the hypotenuse AB as its diameter. The length of the diameter $D$ is the distance between A and B: $D = \sqrt{(a-0)^2 + (0-b)^2} = \sqrt{a^2+b^2}$. The equation of the circle circumscribing $\triangle OAB$ is $x^2 - ax + y^2 - by = 0$. The tangent to the circle at the origin $O(0,0)$ is $ax+by=0$. $d_1$ is the distance of the tangent line $ax+by=0$ from point $A(a,0)$: $d_1 = \frac{|a(a) + b(0) + 0|}{\sqrt{a^2+b^2}} = \frac{a^2}{\sqrt{a^2+b^2}}$. $d_2$ is the distance of the tangent line $ax+by=0$ from point $B(0,b)$: $d_2 = \frac{|a(0) + b(b) + 0|}{\sqrt{a^2+b^2}} = \frac{b^2}{\sqrt{a^2+b^2}}$. Summing $d_1$ and $d_2$: $d_1 + d_2 = \frac{a^2}{\sqrt{a^2+b^2}} + \frac{b^2}{\sqrt{a^2+b^2}} = \frac{a^2+b^2}{\sqrt{a^2+b^2}} = \sqrt{a^2+b^2}$. Since $D = \sqrt{a^2+b^2}$, we have $D = d_1+d_2$.