Let $M\left(\frac{-7}{2}, \frac{-5}{2}\right)$ be the midpoint of the chord $A B$ of the circle $x^2+y^2+10 x+8 y-23=0$. If $a x+$ by $1=0$ is the equation of $A B$ then $3 a+3 b=$

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Aicharya · Accepted Answer

Let $m_1$ and $m_2$ be the slopes of line $O M$ and line $M B$. Hence $m_1=\frac{4-\frac{5}{2}}{5-\frac{7}{2}}=1$ Since $O M \perp M B$ Hence $m_1 m_2=-1$ $ \Rightarrow m_2=-1 $ Hence equation of line $A B$ can be written as, $ \begin{aligned} & \Rightarrow\left(y+\frac{5}{2}\right)=(-1)\left(x+\frac{7}{2}\right) \\ & \Rightarrow x+y+\frac{5}{2}+\frac{7}{2}=0 \\ & \Rightarrow x+y+6=0 \end{aligned} $ $\Rightarrow \frac{1}{6} x+\frac{1}{6} y+1=0$ ...(i) But $a x+b y+1=0$ (given) is also the equation of chord $A B$ Hence $a=\frac{1}{6}, b=\frac{1}{6}$ $\Rightarrow 3 a+3 b=3\left(\frac{1}{6}\right)+3\left(\frac{1}{6}\right)=1$

Let $M\left(\frac{-7}{2}, \frac{-5}{2}\right)$ be the midpoint of the chord $A B$ of the circle $x^2+y^2+10…

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