Question:
<p>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</p>
<p style="display:inline"><span class="math-tex">\(2 \sqrt{5}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{\sqrt{5}}{2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{\sqrt{5}}{4}\)</span></p>
<p style="display:inline"><span class="math-tex">\(4 \sqrt{5}\)</span></p>
Step-by-Step Solution
Key Concept: Determine the circle's equation using the intercepts as the diameter, then find the tangent at the origin using the T=0 rule to calculate the required perpendicular distances.
<p>According to given information, we have the following figure.<br />
<img alt="" data-imgur-src="9IjcEbX.png" src="https://media-mycbseguide.s3.amazonaws.com/images/imgur/9IjcEbX.png" style="width: 200px; height: 162px;" /><br />
From figure, equation of circle (diameter form) is (x - 1) (x - 0) + (y - 0) <span class="math-tex">\(\left(y-\frac{1}{2}\right)\)</span> = 0<br />
<span class="math-tex">\(\Rightarrow\)</span> x<sup>2</sup> + y<sup>2</sup> - x - <span class="math-tex">\(\frac{y}{2}\)</span> = 0<br />
Equation of tangent at (0, 0) is x + <span class="math-tex">\(\frac{y}{2}\)</span> = 0<br />
[<span class="math-tex">\(\because\)</span> equation of tangent at (x<sub>1</sub>, y<sub>1</sub>) is given by T = 0<br />
Here, T = 0<br />
<span class="math-tex">\(\Rightarrow\)</span> xx<sub>1</sub> + yy<sub>1</sub> - <span class="math-tex">\(\frac{1}{2}\)</span> (x + x<sub>1</sub>) - <span class="math-tex">\(\frac{1}{4}\)</span> (y + y<sub>1</sub>) = 0]<br />
<span class="math-tex">\(\Rightarrow\)</span> 2x + y = 0<br />
Now, <span class="math-tex">\(A M=\frac{|2 \cdot 1+1 \cdot 0|}{\sqrt{5}}=\frac{2}{\sqrt{5}}\)</span><br />
[<span class="math-tex">\(\because\)</span> distance of a point P(x<sub>1</sub>, y<sub>1</sub>) from a line ax + by + c = 0 is <span class="math-tex">\(\frac{\left|a x_{1}+b y_{1}+c\right|}{\sqrt{a^{2}+b^{2}}}]\)</span><br />
and <span class="math-tex">\(B N=\frac{\left|2 \cdot 0+1\left(\frac{1}{2}\right)\right|}{\sqrt{5}}=\frac{1}{2 \sqrt{5}}\)</span><br />
<span class="math-tex">\(\therefore\)</span> AM + BN <span class="math-tex">\(=\frac{2}{\sqrt{5}}+\frac{1}{2 \sqrt{5}}=\frac{4+1}{2 \sqrt{5}}=\frac{\sqrt{5}}{2}\)</span></p>
Correct Answer: B