The equation of normal at \((1,1)\) to the circle \(x^2+y^2-x-3 y-4=0\) is
The equation of normal at \((1,1)\) to the circle \(x^2+y^2-x-3 y-4=0\) is
\(x+y-2=0\)
\(2 x-y-1=0\)
\(x-y+2=0\)
\(x-y-2=0\)
Solution
Circle is,
\(x^2+y^2-x-3 y-4=0\)
Slope of tangent at \((1,1)\) is obtained by differentiating above equation,
\(\begin{aligned}
& 2 x+2 y y^{\prime}-1-3 y^{\prime}=0 \\
& \Rightarrow \quad y^{\prime}=\left.\frac{1-2 x}{2 y-3}\right|_{(1,1)} \\
& \Rightarrow y^{\prime}=\frac{1-2}{2-3}=1
\end{aligned}\)
So, slope of normal is,
\(m_N=-1 / m_T=-1\)
Equation of normal in one point forms is,
\(\begin{aligned}
& y-y_1=m\left(x-x_1\right) \\
& \Rightarrow \quad y-1=-1(x-1) \\
& \Rightarrow \quad x+y-2=0 \\
\end{aligned}\)