Question

The angle between the tangents drawn from the point P(k, 6k) to the circle \(x^2+y^2+6x-6y+2=0\) is \(2\tan^{-1}(\frac{4}{3})\). If the coordinates of P are integers, then k =

• A. 1
• B. 2
• C. 3
• D. -2

Hint:

If the angle between the tangents from a point P to a circle is \( \theta \), the half-angle \( \phi = \theta/2 \) is formed by the tangent and the line joining P to the center. In the resulting right-angled triangle, \( \sin\phi = r/d \), \( \tan\phi = r/L \), where r is the radius, d is the distance from P to the center, and L is the length of the tangent (\(\sqrt{S_{11}}\)).

Answer 1

The Correct Option is A

Let \( \theta \) be the angle between the tangents. We are given \( \theta = 2\tan^{-1}(\frac{4}{3}) \).
This means the half-angle, \( \phi = \theta/2 \), is \( \tan^{-1}(\frac{4}{3}) \), so \( \tan\phi = 4/3 \).
The half-angle \( \phi \) is related to the radius of the circle \(r\) and the length of the tangent from P, which is \(L = \sqrt{S_{11}}\).
The relationship is \( \tan\phi = \frac{r}{L} = \frac{r}{\sqrt{S_{11}}} \).
First, find the center and radius of the circle \(x^2+y^2+6x-6y+2=0\).
Center \(C = (-3, 3)\). Radius \(r = \sqrt{(-3)^2 + 3^2 - 2} = \sqrt{9+9-2} = \sqrt{16} = 4\).
Next, find the power of the point P(k, 6k), \(S_{11}\).
\( S_{11} = k^2 + (6k)^2 + 6(k) - 6(6k) + 2 = k^2 + 36k^2 + 6k - 36k + 2 = 37k^2 - 30k + 2 \).
Now, substitute into the formula \( \tan\phi = \frac{r}{\sqrt{S_{11}}} \).
\( \frac{4}{3} = \frac{4}{\sqrt{37k^2 - 30k + 2}} \).
From this, we can see that \( \sqrt{37k^2 - 30k + 2} = 3 \).
Squaring both sides: \( 37k^2 - 30k + 2 = 9 \).
\( 37k^2 - 30k - 7 = 0 \).
We need to find integer solutions for k. We can use the quadratic formula or try to factor.
Let's test integer values. If \(k=1\), \(37(1)^2 - 30(1) - 7 = 37 - 30 - 7 = 0\).
Since \(k=1\) is a solution and it is an integer, this is our required value.
(The other root from the quadratic formula is \(k = -7/37\), which is not an integer).
So, \(k=1\).