Let the circle C touch the line x - y + 1 = 0, have the center on the positive x-axis, and cut off a chord of length 4 1

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Let the circle C touch the line x - y + 1 = 0, have the center on the positive x-axis, and cut off a chord of length 4 13 along the line -3x + 2y = 1. Let H be the hyperbola x^2 ^2 - y^2 ^2 = 1,…

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Given: The circle has center on the positive x-axis, touches the line $x-y+1=0$ , and the chord cut by the line $-3x+2y=1$ has length $\frac{4}{\sqrt{13}}$ .

Find: The value of $2\alpha^2+3\beta^2$ for the hyperbola $\frac{x^2}{\alpha^2}-\frac{y^2}{\beta^2}=1$ .

Let the center of the circle be $(a,0)$ and radius be $r$ . Since the circle touches the line $x-y+1=0$ , the perpendicular distance from $(a,0)$ to this line equals $r$ :

\frac{|a+1|}{\sqrt{2}}=r

As the center lies on the positive x-axis, we take

r=\frac{a+1}{\sqrt{2}}.

Now use the chord-length condition for the line $-3x+2y=1$ . Writing it as $3x-2y+1=0$ , the perpendicular distance from the center $(a,0)$ is

d=\frac{|3a+1|}{\sqrt{13}}.

For a circle, chord length cut by a line at distance $d$ from the center is

2\sqrt{r^2-d^2}.

So,

2\sqrt{r^2-d^2}=\frac{4}{\sqrt{13}}.

Substitute $r=\frac{a+1}{\sqrt{2}}$ and $d=\frac{|3a+1|}{\sqrt{13}}$ :

2\sqrt{\frac{(a+1)^2}{2}-\frac{(3a+1)^2}{13}}=\frac{4}{\sqrt{13}}.

Squaring,

4\left(\frac{(a+1)^2}{2}-\frac{(3a+1)^2}{13}\right)=\frac{16}{13}.

Hence,

26(a+1)^2-4(3a+1)^2=16.

Expanding,

26a^2+52a+26-36a^2-24a-4=16

-10a^2+28a+6=0

5a^2-14a-3=0.

Solving,

a=\frac{14\pm16}{10}

so

a=3 \quad \text{or} \quad a=-\frac{1}{5}.

Since the center lies on the positive x-axis, we take

a=3.

Therefore,

r=\frac{a+1}{\sqrt{2}}=\frac{4}{\sqrt{2}}=2\sqrt{2}.

For the hyperbola $\frac{x^2}{\alpha^2}-\frac{y^2}{\beta^2}=1$ , the transverse axis length is $2\alpha$ . It is given equal to the diameter of the circle, which is $2r$ . Thus,

2\alpha=2r

so

\alpha=r=2\sqrt{2} \Rightarrow \alpha^2=8.

One focus of the hyperbola is the center of the circle, that is at distance $3$ from the origin. For the hyperbola, focus distance $c$ satisfies

c^2=\alpha^2+\beta^2.

Hence,

3^2=\alpha^2+\beta^2.

So,

9=8+\beta^2 \Rightarrow \beta^2=1.

Now compute

2\alpha^2+3\beta^2=2(8)+3(1)=16+3=19.

Therefore, the required value is $19$ .

Given: A circle with center on the positive x-axis touches $x-y+1=0$ and the line $-3x+2y=1$ cuts a chord of length $\frac{4}{\sqrt{13}}$ .

Find: The value of $2\alpha^2+3\beta^2$ for the associated hyperbola.

The tangent condition gives radius as the distance from the center to the tangent line. The chord condition gives a second relation using

\text{chord length}=2\sqrt{r^2-d^2}.

After finding the circle, use the hyperbola facts:

transverse axis length $=2\alpha$ ,
focus distance satisfies $c^2=\alpha^2+\beta^2$ .

The solution also shows an incorrect intermediate possibility $a=-\frac{1}{5}$ in one approach, but that contradicts the condition that the center lies on the positive x-axis. Hence the valid center is $(3,0)$ , which leads consistently to the final answer $19$ .

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