AIEEE 2009 (Offline)Question 1
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 :
Circle
Solution
Concept. Every circle through $P, Q$ is $S_1+\lambda S_2=0$ ($\lambda\ne-1$) or $S_2=0$. For $\lambda=-1$ we get the radical axis, a line. Formulas used. (1) Curves through the common points of $S_1=0, S_2=0$: $S_1+\lambda S_2=0$.Step 1 - Values at $(1,1)$. $S_1(1,1)=1+1+3+7+2p-5=7+2p$. $S_2(1,1)=1+1+2+2-p^2=6-p^2$. Step 2 - Use (1). Need $(7+2p)+\lambda(6-p^2)=0$. If $6-p^2\ne0$, $\lambda=-\frac{7+2p}{6-p^2}$. If $6-p^2=0$, then $S_2$ itself passes through $(1,1)$ (here $7+2p\ne0$). Step 3 - Bad case. The curve is a line (no circle) only when $\lambda=-1$: $$7+2p=6-p^2\Rightarrow p^2+2p+1=0\Rightarrow p=-1.$$ Then $(1,1)$ lies on line $PQ$ and no circle passes through three collinear points. Step 4 - Result. A circle exists for all $p$ except $p=-1$. Answer: A