Question

If the line $3x-4y=m$ cuts the circle $x^2+y^2-4x-8y-5=0$ in two points then limits of $m$ are

• A. $[-35,15]$
• B. $(-35,15)$
• C. $(-35,10)$
• D. $(-35,15]$

Answer 1

Answer: (B)

$(-35,15)$

Given Circle:$x^2+y^2-4x-8y-5=0$

Centre$=(2,4)$

Radius$=\sqrt{4+16+5}=\sqrt{25}=5$

If the circle is intersecting line $3x-4y=m$ at two distinct points.

Length of perpendicular from centre to the line $<$ radius

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

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

$\Rightarrow |10+m|<25$

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

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

$\Rightarrow -35<m<15$

Hence, $m\in (-35,15)$.