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 from tangent to the circle at origin is:
A. $\dfrac{\sqrt{5}}{4}$
B. $\dfrac{\sqrt{5}}{2}$
C. $2\sqrt{5}$
D. $4\sqrt{5}$

Answer 1

We can conclude that AOB is a right angled triangle and angle made by a semi- circle of circle on its circumference is a right angle. Clearly AB is the diameter of the circle and midpoint of AB is the center. Where A is $\left( 0,\dfrac{1}{2} \right)$ and B is $\left( 1,0 \right)$.

Complete step by step answer:
It is given that A and B are on the coordinate axes. Let us consider that A is on the y – axis and B is on the x – axis. This means x – coordinate of A will be 0 and y – coordinate of B will be 0.
Thus, to find the coordinates of A, substitute x = 0 in the line equation $x+2y=1$.
$\begin{aligned} & \Rightarrow \left( 0 \right)+2y=1 \\\ & \Rightarrow y=\dfrac{1}{2} \\\ \end{aligned}$
Thus, coordinates of A are $\left( 0,\dfrac{1}{2} \right)$.
To find B, substitute y = 0 in the equation $x+2y=1$.
$\begin{aligned} & \Rightarrow x+2\left( 0 \right)=1 \\\ & \Rightarrow x=1 \\\ \end{aligned}$
Thus, coordinates of B are $\left( 1,0 \right)$.
We can summarize this question in the form of a figure given below, where AOB is a right angled triangle with right angle at O. And a tangent at origin O on the circle that is required for later part of question,

The $\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}={{d}_{12}}$ where ${{d}_{12}}$ is distance between $\left( {{x}_{1,}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ .This is Distance Formula for 2D Euclidean Space.
As AOB is clearly a right angled triangle and angle made by a semi- circle of circle on its circumference is a right angle. Clearly AB is the diameter of the circle and midpoint of AB is the center. Where A is $\left( 0,\dfrac{1}{2} \right)$ and B is $\left( 1,0 \right)$.
We can use above observation to start the solution of given problem, mid-point of AB is C $C=\left( \dfrac{x\left( A \right)+x\left( B \right)}{2},\dfrac{y\left( A \right)+y\left( B \right)}{2} \right)=\left( \dfrac{0+1}{2},\dfrac{\dfrac{1}{2}+0}{2} \right)=\left( \dfrac{1}{2},\dfrac{1}{4} \right)$ which clearly is center of circle.
The distance between origin and coordinate C gives radius ${{r}_{AB}}=\sqrt{\dfrac{1}{4}+\dfrac{1}{16}}=\sqrt{\dfrac{5}{16}}=\dfrac{\sqrt{5}}{4}$ and the equation of circle is given by $\sqrt{{{\left( x-\dfrac{1}{2} \right)}^{2}}+{{\left( y-\dfrac{1}{4} \right)}^{2}}}=\dfrac{\sqrt{5}}{4}$ which on further simplification.
${{x}^{2}}+{{y}^{2}}-x-\dfrac{y}{2}=0...eq(1)$
For second order conic curve, to find tangent at any point $\left( a,b \right)$ on it we replace
${{x}^{2}}\text{with } ax,{{y}^{2}}\text{with by },x\text{ with }\dfrac{x+a}{2},y\text{ with } \dfrac{y+b}{2}$ in the equation of circle and resultant linear function is tangent at that point. Hence, the equation of line tangent at $\left( 0,0 \right)$ is given as
$0x+0y-\left( \dfrac{x+0}{2} \right)-\dfrac{1}{2}\left( \dfrac{y+0}{2} \right)=0$
$2x+y=0...eq(2)$
If the equation of line is $ax+by+c=0$ and there is a point $\left( h,k \right)$ then distance d given by formula
$d=\dfrac{ah+bk+c}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$ Represents length of perpendicular drawn from point $\left( h,k \right)$ to the above line. Hence distance d1 and d2 are perpendicular distances from points A and B.
Now we try to find perpendicular distance of points A and B from the tangent at origin O, we use formula stated above
${{d}_{1}}=\dfrac{2(0)+1(\dfrac{1}{2})}{\sqrt{{{(2)}^{2}}+{{(1)}^{2}}}}$, ${{d}_{2}}=\dfrac{2(1)+1(0)}{\sqrt{{{(2)}^{2}}+{{(1)}^{2}}}}$ are perpendicular distance from points A and B respectively.
${{d}_{1}}=\dfrac{1}{2\sqrt{5}}$ and ${{d}_{2}}=\dfrac{2}{\sqrt{5}}$ and the sum of distance comes out to be ${{d}_{1}}+{{d}_{2}}=\dfrac{\sqrt{5}}{2}$

So, the correct answer is “Option B”.

Note: The above method to find the equation of tangent applies only to second degree conic curves and for higher degree conic curves calculus is used. The common mistakes committed by the students are incorrect substitution and wrongly writing perpendicular distance formulas.