Intercept made by a line between two concentric circles

Question

Let $$x^2+y^2-9=4r^2\enspace (r=1,2,3)$$ be $3$ concentric cirlces. Prove that the intercept made by line $$3x+4y+15=0$$ between any two cirlces is same.

I thought of calculating the intercept between any two cirlces and then prove it that they are numerically same. But, it would be too long. So, can someone suggest me an elegant method for proving it? Just give me the working lines.

Answer 1

$3(x+5)=-4y\iff\dfrac{x+5}4=\dfrac y{-3}=t$(say)

$\implies x=4t-3,y=-3t$

$$x^2+y^2=4r^2+9\implies(4t-3)^2+(-3t)^2=4r^2+9\iff25t^2-24t-4r^2=0$$

$(t_1-t_2)^2=(t_1+t_2)^2-4t_1t_2=\dfrac{576+400r^2}{625}$

If $(x_1,y_1);(x_2,y_2)$ be the intersections

$$D_r^2=(x_1-y_1)^2+(x_2-y_2)^2=25(t_1-t_2)^2=\dfrac{576+400r^2}{25}$$

Put $r=1,2,3$

Answer 2

Let $L$ cut $C_1$, $C_2$, and $C_3$ at $A$, $B(–5, 0)$, and $C$ respectively. $POJ$ is the line passing through the origin and is normal to line $L$.

$\triangle OJB$ is similar to $\triangle OP’P$. Therefore, $OJ = 3k$, $JB = 4k$ and $OB = 5k$; for some $k$.

$OB = 5$ units implies $k = 1$. Therefore, $OJ = 3$, $JB = 4$ and $OB = 5$.

Applying Pythagoras theorem to $\triangle OJA$, $\triangle OJB$ and $\triangle OJC$, we have $AJ = 2$, $AB = 2$ and then $BC = 2$.