Question:
<p>Let P be the point on the parabola, y<sup>2</sup> = 8x which is at a minimum distance from the centre C of the circle, x<sup>2</sup> + (y + 6)<sup>2</sup> = 1. Then the equation of the circle, passing through C and having its centre at P is:</p>
<p style="display:inline">x<sup>2</sup> + y<sup>2</sup> - 4x + 9y + 18 = 0</p>
<p style="display:inline">x<sup>2</sup> + y<sup>2</sup> - 4x + 8y + 12 = 0</p>
<p style="display:inline">x<sup>2</sup> + y<sup>2</sup> - x + 4y - 12 = 0</p>
<p style="display:inline">x<sup>2</sup> + y<sup>2</sup> - <span class="math-tex">\(\frac{\mathrm{x}}{4}\)</span> + 2y - 24 = 0</p>
Step-by-Step Solution
Key Concept: The shortest distance between a fixed point and a parabola is always measured along the normal to the parabola that passes through that point.
<html><body><p>Minimum distance <span class="math-tex">$\Rightarrow$</span> perpendicular distance<br/>
Eq<sup>n</sup> of normal at p(2t<sup>2</sup>, 4t)<br/>
y = -tx + 4t + 2t<sup>3</sup><br/>
It passes through C(0, -6)<br/>
t<sup>3</sup> + 2t + 3 = 0 <span class="math-tex">$\Rightarrow$</span> t = -1<br/>
<img alt="" data-imgur-src="zax83YN.png" height="130" src="https://media-mycbseguide.s3.amazonaws.com/images/imgur/1621087728-6kfsgu.jpg" width="148"/><br/>
Centre of new circle = P(2t<sup>2</sup>, 4t) = P(2, -4)<br/>
Radius = PC <span class="math-tex">$=\sqrt{(2-0)^{2}+(-4+6)^{2}}=2 \sqrt{2}$</span><br/>
<span class="math-tex">$\therefore$</span> Equation of circle is:<br/>
(x - 2)<sup>2</sup> + (y + 4) <span class="math-tex">$=(2 \sqrt{2})^{2}$</span><br/>
<span class="math-tex">$\Rightarrow$</span> x<sup>2</sup> + y<sup>2</sup> - 4x + 8y + 12 = 0</p></body></html>
Correct Answer: B