If P and Q are the points of intersection of the circles$$ x^2 + y^2 + 3x + 7y +2p – 5= 0$$ and $$x^2 + y^2 +2x + 2y – p^2 = 0$$ then there is a circle passing through P, Q, and (1, 1) for ...
a)all except 2 values of p. b)all values of p. c) all except 1 value of p.
My working: $$s2= x^2 + y^2 + 3x + 7y +2p – 5= 0$$ and $$s1=x^2 + y^2 +2x + 2y – p^2 = 0$$ Equation of required circle is $S1+kS2=0$ (family of circles) Putting (1,1), I got $$k= \frac{2p+7}{p^2-6}.$$
Therefore, p is not equal to +/- root 6, i.e. excepting two values of p.
However, by using another method, I'm getting a different answer. $$s2=x^2 + y^2 + 3x + 7y +2p – 5$$ and $$s1=x^2 + y^2 +2x + 2y – p^2 = 0$$ Equation of $PQ= S1-S2.$
Equation of line$ L(PQ)=x+5y+p^2+2p-5.$ Required equation= $S1+ kL=0.$
$$ k= \frac{-2p-7}{(p+1)^2}.$$ Therefore, p is not equal to -1, i.e. excepting one value of p.
The discrepancy arises from the way that you’ve parameterized the family of circles in your first approach. There’s no value of $k$ for which the combination $S_1+kS_2$ produces $S_2$ itself, which is fine as long as $S_2$ doesn’t pass through $(1,1)$, but it in fact does so when $p=\pm\sqrt6$. That’s the true significance of the values of $p$ that you found there.
You might try the combination $kS_1+S_2$ instead, but that suffers from the same problem: $S_1$ passes through $(1,1)$ when $p=-7/2$. You need to use the affine combination $(1-k)S_1+kS_2$ to capture all of the possibilities. This is in fact the same as your second method: $S_1+kL = S_1+k(S_2-S_1)=(1-k)S_1+kS_2$. (Contrary to what you wrote, you actually used $S_2-S_1$, not $S_1-S_2$.) Note that this set of affine combination doesn’t include $S_1-S_2$, but that’s a line, so we’re not interested in it.
It’s worth examining the case $p=-1$ to see why it’s special. For that value of $p$, the radical axis $L$ has the equation $x+5y=6$, which means that it passes through $(1,1)$. There is a linear combination of the two circles that passes through $(1,1)$, but it’s not a circle (unless you consider the line to be a circle with infinite radius).