3. Determine the parameter $\lambda$ such that the distance from the center of the circle

$$
x^{2}+y^{2}+4 x-4 y-17=0
$$

to the line $x-(\lambda+2) y-\lambda-4=0$ is equal to $\frac{5}{\sqrt{2}}$. The tangents that touch the circle at the intersection points with the given line and the line itself form a triangle. Determine the area of the intersection of the circumcircle of that triangle and the given circle. (Take the integer solution for $\lambda$.)
Solution. The equation of the circle $k$ can be written as

$$
(x+2)^{2}+(y-2)^{2}=25
$$

The center of the circle is the point $O(-2,2)$. The equation

$$
\frac{|-2-2(\lambda+2)-\lambda-4|}{\sqrt{1+(\lambda+2)^{2}}}=\frac{5}{\sqrt{2}}
$$

has two solutions $\lambda_{1}=5$ and $\lambda_{2}=-\frac{15}{7}$. The line $x-7 y-9=0$ intersects the circle $k$ at points $A(2,-1)$ and $B(-5,-2)$, see the diagram. The equations of the tangents to the circle $k$ at points $A$ and $B$, respectively, are:

$$
y+1=\frac{4}{3}(x-2) \text { and } y+2=-\frac{3}{4}(x+5)
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_53a7c72cd2352523c04ag-4.jpg?height=625&width=813&top_left_y=312&top_left_x=478)

The intersection point of these tangents is $C(-1,-5)$. Note that the quadrilateral $A O B C$ is a square with side 5. The required area is equal to

$$
\frac{5^{2} \pi}{4}+\frac{1}{2}\left(\frac{50}{4} \pi-25\right)=\frac{25}{2}(\pi-1)
$$