Question

The angle between the tangents drawn from the point \((1,1)\) to the circle \(2x^{2}+2y^{2}-8x+4y+5=0\) is …………

• A. \(\frac{\pi }{2}\)
• B. \(\frac{\pi }{3}\)
• C. \(\frac{\pi }{4}\)
• D. \(\frac{\pi }{6}\)

Hint:

Compare the length of the tangent with the radius, and then use tan of half the angle equals r over L.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Tangent length from a point P is \(L=\sqrt{S_{1}}\), and the angle \(2\phi\) between tangents satisfies \(\tan\phi=\dfrac{r}{L}\).

Step 2: Circle Details
Divide by 2: \(x^{2}+y^{2}-4x+2y+\dfrac{5}{2}=0\).
Centre \((2,-1)\) and \(r^{2}=4+1-\dfrac{5}{2}=\dfrac{5}{2}\).

Step 3: Tangent Length
\(L^{2}=1+1-4+2+\dfrac{5}{2}=\dfrac{5}{2}\).

Step 4: Angle
\(L=r\), so \(\tan\phi=1\) and \(\phi=\dfrac{\pi}{4}\).
The angle between the tangents is \(2\phi=\dfrac{\pi}{2}\).

Final Answer:
The angle between the tangents is \(\dfrac{\pi}{2}\), option (A). \[ \boxed{\text{(A) }\dfrac{\pi }{2}} \]