Mathematics - Circle Question with Solution | TestHub

Circle $x^2+y^2-4x-8y-5=0$ Centre $=(2,4)$, Radius $=\sqrt{4+16+5}=5$ If circle is intersecting line $3x-4y=m$ at two distinct points. $\Rightarrow$ length of perpendicular from centre $<$ radius $\Rightarrow \frac{|6-16-\mathrm{m}|}{5}<5$ $\Rightarrow |10+\mathrm{m}|<25$ $\Rightarrow -25<m+10<25$ $\Rightarrow -35<\mathrm{m}<15$.