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If the circles $x^2+y^2+2 a x+c y+a=0$ and $x^2+y^2-3 a x+d y-1=0$ intersect in two distinct points $P$ and $Q$ then the line $5 x+$ by $-a=0$ passes through $P$ and $Q$ for
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1815 Upvotes
Verified Answer
The correct answer is:
no value of a
no value of a
$$
\begin{aligned}
& S_1=x^2+y^2+2 a x+c y+a=0 \\
& S_2=x^2+y^2-3 a x+d y-1=0
\end{aligned}
$$
Equation of radical axis of $S_1$ and $S_2$
$$
\begin{aligned}
& S_1-S_2=0 \\
& \Rightarrow 5 a x+(c-d) y+a+1=0
\end{aligned}
$$
Given that $5 \mathrm{x}+\mathrm{by}-\mathrm{a}=0$ passes through $\mathrm{P}$ and $\mathrm{Q}$
$$
\begin{aligned}
& \Rightarrow \frac{a}{1}=\frac{c-d}{b}=\frac{a+1}{-a} \\
& \Rightarrow a+1=-a^2 \\
& a^2+a+1=0
\end{aligned}
$$
No real value of a.
\begin{aligned}
& S_1=x^2+y^2+2 a x+c y+a=0 \\
& S_2=x^2+y^2-3 a x+d y-1=0
\end{aligned}
$$
Equation of radical axis of $S_1$ and $S_2$
$$
\begin{aligned}
& S_1-S_2=0 \\
& \Rightarrow 5 a x+(c-d) y+a+1=0
\end{aligned}
$$
Given that $5 \mathrm{x}+\mathrm{by}-\mathrm{a}=0$ passes through $\mathrm{P}$ and $\mathrm{Q}$
$$
\begin{aligned}
& \Rightarrow \frac{a}{1}=\frac{c-d}{b}=\frac{a+1}{-a} \\
& \Rightarrow a+1=-a^2 \\
& a^2+a+1=0
\end{aligned}
$$
No real value of a.
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