Let the circle C touch the line x-y+1=0, have its centre on the positive x-axis, and cut o…

Question

Let the circle $C$ touch the line $x-y+1=0$, have its centre on the positive $x$-axis, and cut off a chord of length $\dfrac{4}{\sqrt{13}}$ along the line $-3x+2y=1$. Let $H$ be the hyperbola $\dfrac{x^2}{\alpha^2}-\dfrac{y^2}{\beta^2}=1$, whose one focus is the centre of $C$ and the length of the transverse axis is the diameter of $C$. Then $2\alpha^2+3\beta^2$ is equal to \_\_\_\_\_\_\_\_.

Answer: 19

Accepted Answer

Let centre of $C$ be $(a,0)$, radius $r$. Tangency: $\dfrac{|a+1|}{\sqrt2}=r\Rightarrow (a+1)^2=2r^2$ (1).

Half-chord on $-3x+2y-1=0$: $\left(\dfrac{2}{\sqrt{13}}\right)^2+d^2=r^2$, with $d=\dfrac{|3a+1|}{\sqrt{13}}$:

$$(3a+1)^2+4=13r^2 \quad(2).$$

Combining (1),(2): $5a^2-14a-3=0 \Rightarrow (5a+1)(a-3)=0$. Since centre is on the positive $x$-axis, $a=3$, and from (1): $r^2=8\Rightarrow r=2\sqrt2$.

For $H$: focus at $(3,0)\Rightarrow \alpha e=3$; transverse axis $2\alpha=2r=4\sqrt2\Rightarrow \alpha=2\sqrt2,\ \alpha^2=8$.

$$\alpha^2e^2=9 \;(\text{automatically satisfied}),\qquad \beta^2=\alpha^2(e^2-1)=\alpha^2e^2-\alpha^2=9-8=1.$$

$$2\alpha^2+3\beta^2=16+3=19.$$

Solutions · 1