JEE Main 2023 (Online) 13th April Evening ShiftQuestion 1
Let the centre of a circle $C$ be $(\alpha,\beta)$ and its radius $r<8$. Let $3x+4y=24$ and $3x-4y=32$ be two tangents and $4x+3y=1$ be a normal to $C$. Then $(\alpha-\beta+r)$ is equal to
Circle
Solution
Normal passes through centre: $4\alpha+3\beta=1$. Distances to tangents equal $r$: $|3\alpha+4\beta-24|=5r$, $|3\alpha-4\beta-32|=5r$. Solving with $r<8$ gives $\alpha=1,\beta=-1,r=5$. $\alpha-\beta+r=1+1+5=7$. Option (C).