Need help in geometry, please!

1. 

(a) The line 𝑥 + 2𝑦 = 0 meets the circle x² + y² − 4𝑥 + 2𝑦 = 0 at A and B. Find the length of AB.

\(\begin{array}{|rclrcl|} \hline x+2y &=& 0 \\ x &=& -2y \\\\ x^2 + y^2 - 4x + 2y &=& 0 \quad & | \quad x=-2y \\ (-2y)^2 + y^2 - 4(-2y) + 2y &=& 0 \\ 4y^2 + y^2 + 8y + 2y &=& 0 \\ 5y^2 + 10y &=& 0 \\ 5y(y+2) &=& 0 \\\\ 5y &=& 0 & y+2 &=& 0 \\ y &=& 0 & y &=& -2 \\\\ x &=& -2\cdot 0 & x&=& -2(-2) \\ x &=& 0 & x&=& 4 \\\\ & & A(0,0) &&& B(4,-2) \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline AB &=& \sqrt{(0-4)^2+(0-(-2))^2} \\ &=& \sqrt{16+(0+2)^2} \\ &=& \sqrt{16+4} \\ &=& \sqrt{20} \\ &=& \sqrt{4\cdot 5} \\ &=& 2\sqrt{ 5} \\ \hline \end{array}\)

(b) The circle 𝑥2 + 𝑦2 − 4𝑥 − 6𝑦 + 𝑘 = 0 touches the x-axis. Find k and the
coordinates of the point of contact. Find the coordinates of the centre and the radius of the circle.

\(k=\ ?\)

\(\begin{array}{|rcll|} \hline x^2+y^2-4x-6y +k &=& 0 \quad & | \quad y = 0 (\text{ circle touches the x-axis} ) \\ x^2 -4x +k &=& 0 \\ x &=& \frac{4\pm \sqrt{16-4k} }{2} \quad & | \quad \text{ only one solution for x} \\ 16-4k &=& 0 \\ 4k &=& 16 \\ k &=& 4 \\\\ x &=& \frac{4\pm 0 }{2}\\ x &=& 2 \\ \hline \end{array} \)

\( k = 4\\\text{The coordinates of the point of contact is $ (2,0)$}\)

\(\text{Coordinates of the centre $(c_x, cr_y)$ and the radius $r$ of the circle.} \)
\(\begin{array}{|rcll|} \hline x^2+y^2-4x-6y +k &=& 0 \\ x^2-4x+y^2-6y +k &=& 0 \\ (x-2)^2-4+(y-3)^2 -9 +k &=& 0 \\ (x-2)^2 +(y-3)^2 &=& 4+9-k \quad & | \quad k = 4 \\ (x-2)^2 +(y-3)^2 &=& 4+9-4 \\ (x-2)^2 +(y-3)^2 &=& 9 \\ (x-\underbrace{2}_{=\text{c_x}})^2+(y-\underbrace{3}_{=\text{c_y}})^2 &=& \underbrace{9}_{=r^2} \\\\ c_x &=& 2 \\ c_y &=& 3 \\ r &=& \sqrt{9} = 3 \\ \hline \end{array}\)

\(\text{The coordinates of the centre $(2,3)$} \\ \text{Radius of the circle $r = 3$ } \)

Thank you so much, you're awsome :)