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 :

Circle PYQ — JEE Main 2019 · 11 Jan · Shift 1 · Q20 with Solution

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Correct answer: B

The line is $x+2y=1.$

On the $x$ -axis, $y=0$ : $x=1 \Rightarrow A=(1,0).$
On the $y$ -axis, $x=0$ : $2y=1 \Rightarrow y=\frac12 \Rightarrow B=\left(0,\frac12\right).$

The circle passes through $O=(0,0)$ , $A=(1,0)$ , and $B=\left(0,\frac12\right)$ .

General equation: $x^2+y^2+2gx+2fy+c=0.$

Since it passes through the origin, $c=0$ . So $x^2+y^2+2gx+2fy=0.$

Now substitute $A=(1,0)$ : $1+2g=0 \Rightarrow g=-\frac12.$

Substitute $B=\left(0,\frac12\right)$ :

\Rightarrow \frac14+f=0 \Rightarrow f=-\frac14.$$ Hence the circle is $$x^2+y^2-x-\frac y2=0.$$ --- 3. **Find tangent at the origin** For the circle $$x^2+y^2+2gx+2fy=0,$$ the tangent at the origin is $$gx+fy=0.$$ Here, $$g=-\frac12, \qquad f=-\frac14.$$ So the tangent is $$-\frac12 x-\frac14 y=0.$$ Multiplying by $-4$, $$2x+y=0.$$ --- 4. **Find perpendicular distances of $A$ and $B$ from the tangent** The tangent line is $$2x+y=0.$$ Distance of point $(x_1,y_1)$ from line $2x+y=0$ is $$d=\frac{|2x_1+y_1|}{\sqrt{2^2+1^2}}=\frac{|2x_1+y_1|}{\sqrt5}.$$ ### Distance of $A=(1,0)$: $$d_A=\frac{|2(1)+0|}{\sqrt5}=\frac{2}{\sqrt5}.$$ ### Distance of $B=\left(0,\frac12\right)$: $$d_B=\frac{\left|0+\frac12\right|}{\sqrt5}=\frac{1}{2\sqrt5}.$$ --- 5. **Sum of distances** $$d_A+d_B=\frac{2}{\sqrt5}+\frac{1}{2\sqrt5} =\frac{4+1}{2\sqrt5} =\frac{5}{2\sqrt5} =\frac{\sqrt5}{2}.$$ --- 6. **Check with options** $$\frac{\sqrt5}{2}$$ matches **Option B**. --- 7. **Comparison with stored correct answer** Stored correct answer: **B** Our derived answer: **B** So they agree.