Q.The line $2x-y+6=0$ meets the circle $x^2+y^2+10x+9=0$ at $A$ and $B$. Find the equation of the circle on $AB$ as diameter.

Any circle through the two intersection points $A,B$ of the given line and circle can be written as $x^2+y^2+10x+9+\lambda(2x-y+6)=0$, i.e. $x^2+y^2+(10+2\lambda)x-\lambda y+(9+6\lambda)=0$, for some constant $\lambda$. For $AB$ to be a diameter of this new circle, its centre $\left(-\dfrac{10+2\lambda}{2},\ \dfrac{\lambda}{2}\right)=(-(5+\lambda),\ \lambda/2)$ must itself lie on the line $2x-y+6=0$ (since the centre of a circle always lies on any diameter): $2(-(5+\lambda))-\dfrac{\lambda}{2}+6=0 \Rightarrow -10-2\lambda-\dfrac{\lambda}2+6=0 \Rightarrow -4-\dfrac52\lambda=0 \Rightarrow \lambda=-\dfrac85$. Substituting back: $10+2\lambda=10-\dfra …

Family of circles through the line-circle intersection: circle $+\lambda\cdot$line$=0$; force the new circle's centre to lie on the given line (since a circle …

• Forgetting to actually solve for the two intersection points if attempting the direct route instead, and mixing up which quantity ($AM=BM$ vs full $AB$) is used …