Question

The length of the tangent segment drawn from point \(P\) to the circle, \(x^{2}+y^{2}-2x-2y-2=0\) is \(4\sqrt{2}\) units. If \(\lambda _{max}\) and \(\lambda _{min}\) are the maximum and minimum distances of point \(P\) from the circle, then \(\lambda _{max}+\lambda _{min}\) is equal to....

• A. \(6\)
• B. \(8\)
• C. \(12\)
• D. \(16\)

Hint:

Find the distance of P from the centre using the tangent length, then add the maximum and minimum distances.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
For a point P at distance d from the centre of a circle of radius r, the maximum distance is \(d+r\) and the minimum is \(d-r\).

Step 2: Circle data
\(x^{2}+y^{2}-2x-2y-2=0\) has centre \((1,1)\) and radius \(r=\sqrt{1+1+2}=2\).

Step 3: Find d
Tangent length: \(\ell^{2}=d^{2}-r^{2}\Rightarrow 32=d^{2}-4\Rightarrow d=6\).

Step 4: Add
\(\lambda_{max}+\lambda_{min}=(6+2)+(6-2)=12\)

Final Answer:
The required sum is 12. \[ \boxed{\text{(C) }12} \]