Question

If $3x+4y+\lambda -3=0$ and $3x+4y+\lambda +3=0$ are the chords of $x^2+y^2+6x+10y+30=0$, then number of integral value(s) of $\lambda$ is

Answer 1

Given lines,
$L_1:3x+4y+\lambda -3=0,$
$L_2:3x+4y+\lambda +3=0$
and circle
$x^2+y^2+6x+10y+30=0$
$C\equiv (-3,-5)r=\sqrt{(-3{)}^2+(-5{)}^2-30}=2$
The perpendicular distance from the centre of the circle to the line $L_1$ is
$p_1=\frac{|3\times -3+4\times -5+\lambda -3|}{\sqrt{3^2+4^2}}\Rightarrow p_1=\frac{|\lambda -32|}{5}$
As this is chord, so
$p_1<r\Rightarrow \frac{|\lambda -32|}{5}<2\Rightarrow |\lambda -32|<10\Rightarrow -10<\lambda -32<10\Rightarrow 22<\lambda <42\cdots \left(1\right)$
The perpendicular distance from the centre of the circle to the line $L_2$ is
$p_2=\frac{|3\times -3+4\times -5+\lambda +3|}{\sqrt{3^2+4^2}}\Rightarrow p_2=\frac{|\lambda -26|}{5}$
This is a chord, so
$p_2<r\Rightarrow \frac{|\lambda -26|}{5}<2\Rightarrow |\lambda -26|<10\Rightarrow -10<\lambda -26<10\Rightarrow 16<\lambda <36\cdots \left(2\right)$
From equation $\left(1\right)$ and $2$, we get
$\Rightarrow 22<\lambda <36$

Hence, the total number of integral values is $13.$