Question

Circle $x^2+y^2-4x-8y-5=0$ will intersect the line $3x-4y=m$ in two distinct points, if

• A. $-10<m<5$
• B. $9<m<20$
• C. $-35<m<15$
• D. None of these

Answer 1

The correct option is C $-35<m<15$

Given circle equation $x^2+y^2-4x-8y-5=0$

centre = (2,4) radius $=\sqrt{2^2+4^2+5}$

$=5$

A Given equation of line $\Rightarrow 3x-4y=m$

A line will intersect a circle at

two points only if perpendicular distance

of line from centre of line circle is

less than radius

i.e $\left|\frac{3\left(2\right)-4\left(4\right)-m}{\sqrt{3^2+4^2}}\right|<5$

$\left|\frac{6-16-m}{5}\right|<5$

$-5<\frac{-10-m}{5}<5$

$\Rightarrow -25<-10-m<25$

$\Rightarrow -15<-m<35$ [multiply with -1]

$\Rightarrow -35<m<15$