Question:
<p>If P and Q are the points of intersection of the circles x<sup>2</sup> + y<sup>2</sup> + 3x + 7y + 2p - 5 = 0 and x<sup>2</sup> + y<sup>2</sup> + 2x + 2y - p<sup>2</sup> = 0, then there is a circle passing through P, Q and (1, 1) for</p>
<p style="display:inline">all except two values of p</p>
<p style="display:inline">exactly one value of p</p>
<p style="display:inline">all except one value of p</p>
<p style="display:inline">all values of p</p>
Step-by-Step Solution
Key Concept: Use the family of circles passing through intersection of two given circles: S₁ + λL = 0, where L is the radical axis (difference of circle equations). Substitute point (1,1) into this family to find the value(s) of p for which a circle through P, Q, and (1,1) exists.
<p>Equation of circle passing through the intersection of given circles is<br />
(x<sup>2</sup> + y<sup>2</sup> + 3x + 7y + 2p - 5) + <span class="math-tex">$\lambda$</span>(x + 5y + 2p - 5 + p<sup>2</sup>) = 0<br />
Since it passes through (1, 1).<br />
<span class="math-tex">$\therefore$</span> (7 + 2p) + <span class="math-tex">$\lambda$</span>(1 + p)<sup>2</sup> = 0<br />
<span class="math-tex">$\Rightarrow \lambda=-\frac{7+2 p}{(1+p)^{2}}$</span><br />
If p = -1, <span class="math-tex">$\lambda$</span> does not exist.</p>
Correct Answer: C