JEE Main 2019 (Online) 10th April Morning ShiftQuestion 1
The line $x=y$ touches a circle at the point $(1,1)$. If the circle also passes through the point $(1,-3)$, then its radius is
Circle
Solution
Formulas used. (1) The centre lies on the normal at the point of contact (the line through that point perpendicular to the tangent).(2) Radius = distance from the centre to any point of the circle.Step 1 - Normal at $(1,1)$. The tangent $x=y$ has slope $1$, so the normal has slope $-1$: $$y-1=-(x-1)\Rightarrow x+y=2$$ By (1), the centre is $(h,\,2-h)$. Step 2 - Equal distances. The centre is equally far from $(1,1)$ and $(1,-3)$: $$(h-1)^2+(1-h)^2=(h-1)^2+(5-h)^2$$ $$\Rightarrow (1-h)^2=(5-h)^2\Rightarrow 1-2h=25-10h\Rightarrow h=3$$ Step 3 - Centre and radius. Centre $=(3,-1)$. By (2): $$r=\sqrt{(3-1)^2+(-1-1)^2}=\sqrt8=2\sqrt2$$ (Check with $(1,-3)$: $\sqrt{4+4}=2\sqrt2$ ✓.) Answer: D