Question

The straight line $x+2y=1$ meets the coordinate axes at $A$ and $B$. A circle is drawn through $A,B$ and the origin. Then the sum of perpendicular distances from $A$ and $B$ on the tangent to the circle at the origin is :

• A. $4\sqrt{5}$
• B. $\frac{\sqrt{5}}{4}$
• C. $2\sqrt{5}$
• D. $\frac{\sqrt{5}}{2}$

Answer 1

The correct option is D $\frac{\sqrt{5}}{2}$


Given line $AB$ is $x+2y=1$
Slope of line $AB=-\frac{1}{2}$
Co-ordinates of $A$ and $B$ are $A=(1,0),B=(0,\frac{1}{2})$
Co-ordinates of $C=(\frac{1+0}{2},\frac{0+\frac{1}{2}}{2})=(\frac{1}{2},\frac{1}{4})$
Slope of $OC=\frac{0-\frac{1}{4}}{0-\frac{1}{2}}=\frac{1}{2}$

$\therefore$ Slope of tangent $=-2$
So, the equation of tangent is $2x+y=0$
Sum of distances from $A(1,0)$ and $B(0,\frac{1}{2})$ of tangent
$=\left|\frac{2}{\sqrt{2^2+1^2}}\right|+\left|\frac{\frac{1}{2}}{\sqrt{2^2+1^2}}\right|=\frac{2}{\sqrt{5}}+\frac{\frac{1}{2}}{\sqrt{5}}=\frac{5}{2\sqrt{5}}=\frac{\sqrt{5}}{2}$