Question

A line meets the co-ordinate axes an $A$ and $B$. A circle is circumscribed about the $∆ABO$.

If $p$ and $q$ are the distances of the tangents to the circle at the origin from points $A$ and $B$ respectively, then the diameter of the circle is

• A. $q(p+q)$
• B. $p+q$
• C. $p(p+q)$
• D. $p(p-q)$

Answer 1

The correct option is B

$p+q$

Step 1: Draw the figure with the given data:

Let the coordinates of $A$ be $(a,0)$ and the coordinates of $B$ be $(0,b)$

Step 2: Calculating the radius of circle

$\therefore \angle AOB=90°$

$\therefore AB$ is the diameter of the circle.

The center of the circle $C$ has the coordinates $\left(\frac{a}{2},\frac{b}{2}\right)$

The radius of the circle $=\frac{1}{2}\sqrt{a^2+b^2}$

Step 3: Finding the diameter

The equation of the circle is $\begin{array}{rcl}{x}^2+y^2-ax-by& =& 0\end{array}$

The tangent of the circle is $\begin{array}{rcl}ax+by& =& 0\end{array}$

Now,

$\begin{array}{rcl}p& =& \frac{a^2}{\sqrt{a^2+b^2}}\\ q& =& \frac{b^2}{\sqrt{a^2+b^2}}\\ p+q& =& \frac{a^2}{\sqrt{a^2+b^2}}+\frac{b^2}{\sqrt{a^2+b^2}}\\ & =& \frac{a^2+b^2}{\sqrt{a^2+b^2}}\\ & =& \sqrt{a^2+b^2}\end{array}$

$\therefore$The diameter of the circle is $\begin{array}{rcl}\sqrt{{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}}\mathbf{}& \mathbf{=}& \mathbf{}\mathit{p}\mathbf{+}\mathit{q}\end{array}$

Hence, the correct option is (B).