Question

If the angle between the pair of tangents drawn to the circle $ x^2 + y^2 - 2x + 4y + 3 = 0 $ from the point $(6, -5)$ is \(\theta\) than \(\cot \theta\) = 

• A. \(\frac{15}{8}\)
• B. \(\frac{8}{15}\)
• C. \(\frac{7}{5}\)
• D. \(\frac{5}{7}\)

Answer 1

The Correct Option is A

To find the cotangent of the angle \(\theta\) between the pair of tangents from the point \((6, -5)\) to the circle \(x^2 + y^2 - 2x + 4y + 3 = 0\), we proceed as follows:

1. Finding the Center and Radius of the Circle:
The given circle equation is \(x^2 + y^2 - 2x + 4y + 3 = 0\). 
Rewrite it in standard form by completing the square:

For \(x\): \(x^2 - 2x = (x - 1)^2 - 1\)
For \(y\): \(y^2 + 4y = (y + 2)^2 - 4\)
So:

\( (x - 1)^2 - 1 + (y + 2)^2 - 4 + 3 = 0 \)
\( (x - 1)^2 + (y + 2)^2 - 2 = 0 \)
\( (x - 1)^2 + (y + 2)^2 = 2 \)
The center is \((1, -2)\), and the radius is:

\( r = \sqrt{2} \)

2. Distance from Point to Center:
Calculate the distance \(d\) from the point \((6, -5)\) to the center \((1, -2)\):

\( d = \sqrt{(6 - 1)^2 + (-5 - (-2))^2} = \sqrt{5^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34} \)

3. Angle Between Tangents:
Let \(\theta\) be the angle between the pair of tangents from \((6, -5)\) to the circle. 
The relationship between the radius \(r\), distance \(d\), and the angle \(\theta\) is:

\( \sin\left(\frac{\theta}{2}\right) = \frac{r}{d} = \frac{\sqrt{2}}{\sqrt{34}} = \sqrt{\frac{2}{34}} = \sqrt{\frac{1}{17}} = \frac{1}{\sqrt{17}} \)

4. Calculating \(\cos\left(\frac{\theta}{2}\right)\):
Using the identity \(\cos^2\left(\frac{\theta}{2}\right) = 1 - \sin^2\left(\frac{\theta}{2}\right)\):

\( \cos^2\left(\frac{\theta}{2}\right) = 1 - \frac{1}{17} = \frac{16}{17} \)
\( \cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{16}{17}} = \frac{4}{\sqrt{17}} \)

5. Computing \(\cot\theta\):
Use the double-angle identities to find \(\cot\theta\):

\( \cot\theta = \frac{\cos\theta}{\sin\theta} = \frac{\cos^2\left(\frac{\theta}{2}\right) - \sin^2\left(\frac{\theta}{2}\right)}{2 \sin\left(\frac{\theta}{2}\right) \cos\left(\frac{\theta}{2}\right)} \)
Substitute:

\( \cos^2\left(\frac{\theta}{2}\right) = \frac{16}{17}, \quad \sin^2\left(\frac{\theta}{2}\right) = \frac{1}{17} \)
\( \sin\left(\frac{\theta}{2}\right) = \frac{1}{\sqrt{17}}, \quad \cos\left(\frac{\theta}{2}\right) = \frac{4}{\sqrt{17}} \)
So:

\( \cot\theta = \frac{\frac{16}{17} - \frac{1}{17}}{2 \cdot \frac{1}{\sqrt{17}} \cdot \frac{4}{\sqrt{17}}} = \frac{\frac{15}{17}}{\frac{8}{17}} = \frac{15}{17} \cdot \frac{17}{8} = \frac{15}{8} \)

Final Answer:
The cotangent of the angle between the tangents is \(\frac{15}{8}\).