10.1 circle and system of circle — ગણિત ધોરણ 11 સાયન્સ — Question

$x=\frac{4 y}{3}$

$x^2+y^2=25$

$x^2-\left(1+\frac{16}{9}\right)=25$

$x^2=9$

$x=\pm 3$

$y==9 / 4$

Now, $x^2+y^2=25$

$2 x+2 y \frac{d y}{d x}=0$

$\frac{d y}{d x}=\frac{-x}{y}=\frac{-(3)}{(9 / 4)}=-4 / 3$

$\frac{-d x}{d y}=3 / 4 \Rightarrow$ slope of normal at circle and $3 / 4 \Rightarrow$ solpe of $3 x-4 y=0$

So, Line is normal at circle.