Circles
Grade 11

Question:

<p>Let the line segment joining the centres of the circles x<sup>2</sup> - 2x + y<sup>2</sup> = 0 and x<sup>2</sup> + y<sup>2</sup> + 4x + 8y + 16 = 0 intersect the circles at P and Q respectively. Then the equation of the circle with PQ as its diameter is</p>
<p style="display:inline">5x<sup>2</sup> + 5y<sup>2</sup> - 2x - 16y + 8 = 0</p>
<p style="display:inline">5x<sup>2</sup> + 5y<sup>2</sup> + 8x - 24y + 27 = 0</p>
<p style="display:inline">5x<sup>2</sup> + 5y<sup>2</sup> - 8x - 24y + 27 = 0</p>
<p style="display:inline">5x<sup>2</sup> + 5y<sup>2</sup> + 2x +  16y + 8 = 0</p>

Step-by-Step Solution

Key Concept: Identify the center of the diameter PQ by applying the section formula to the line segment joining the centers of the two given circles, using a ratio derived from their radii.
<p>The centres of two circles are C<sub>1</sub>(1, 0) and C<sub>2</sub>(-2, -4) and their radii are 1 and 2 units respectively.<br /> Let C be the centre of the required circle.<br /> Then, CP = CQ = 1.<br /> <span class="math-tex">$\therefore$</span> CC<sub>1</sub> = 2 and CC<sub>2</sub> = 3.<br /> Clearly, C divides C<sub>1</sub> C<sub>2</sub> in the ratio 2 : 3.<br /> Therefore, coordinates of C are<br /> <span class="math-tex">$\left(\frac{-4+3}{2+3}, \frac{-8+0}{2+3}\right)=\left(-\frac{1}{5},-\frac{8}{5}\right)$</span>.<br /> Hence, the equation of the required circle is<br /> <span class="math-tex">$\left(x+\frac{1}{5}\right)^{2}+\left(y+\frac{8}{5}\right)^{2}=1^{2}$</span><br /> <span class="math-tex">$\Rightarrow$</span> 5x<sup>2</sup> + 5y<sup>2</sup> + 2x +&nbsp;16y + 8 = 0<br /> <img alt="" src="https://media-mycbseguide.s3.amazonaws.com/images/imgur/1623761622-hkc25y.jpg" style="height:147px; width:170px" /></p>
Correct Answer: D

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