Band 3 questions 13 Peter's homework was to make an accurate drawing of a triangle with sides \( 5 \mathrm{~cm}, 5 \mathrm{~cm} \) and an interior angle of \( 60^{\circ} \). Peter said I will need to draw two different triangles as I don't know if the angle of \( 60^{\circ} \) is between the two 5 cm sides.

a. Determine the radius and the co-ordinates of the centre of the circle given by the equation \[ x^{2}+y^{2}+8 x-2 y+8=0 \] [5 marks] b. Find the point of intersection of the circle \( x^{2}+y^{2}-2 x-8 y-6=0 \) and the line \( x+y=2 \) c. Find the equation of the tangent and the length of the tangent to the circle \( 4 x^{2}+4 y^{2}-2 x+5 y-8=0 \) from the point \( (-4,4) \). 5 marks \( ] \) d. Given the circle \( (x-2)^{2}+(y+3)^{2}=16 \), sketch the graph of the circle. [5 marks]

১৯. সামান্তরিকের ক্ষেত্রফল নির্ণয়ের সূত্রটি লেখ।

7. En cada ítem, estudiar la posición relativa entre las dos rectas dadas y graficar ambas en un mismo sistema de coordenadas cartesianas. a) \( \left\{\begin{array}{l}x=1-5 \lambda \\ y=2+3 \lambda \\ z=-5+\lambda\end{array}\left\{\begin{array}{l}x=1 \\ y=1 \\ z=\mu\end{array}(\lambda, \mu \in \mathbb{R})\right.\right. \)

7. En cada ítem, estudiar la posición relativa entre las dos rectas dadas y graficar ambas en un mismo sistema de coordenadas cartesianas. a) \( \left\{\begin{array}{l}x=1-5 \lambda \\ y=2+3 \lambda \\ z=-5+\lambda\end{array}\left\{\begin{array}{l}x=1 \\ y=1 \\ z=\mu\end{array}(\lambda, \mu \in \mathbb{R})\right.\right. \)\( \quad \) b) \( \left\{\begin{array}{l}x=3+2 \alpha \\ y=1-\alpha \\ z=5\end{array} \quad\left\{\begin{array}{l}x=-1-6 \beta \\ y=3+3 \beta \\ z=5\end{array} \quad(\alpha, \beta \in \mathbb{R})\right\}\right. \)

7. En cada ítem, estudiar la posición relativa entre las dos rectas dadas y graficar ambas en un mismo sistema de coordenadas cartesianas. a) \( \left\{\begin{array}{l}x=1-5 \lambda \\ y=2+3 \lambda \\ z=-5+\lambda\end{array}\left\{\begin{array}{l}x=1 \\ y=1 \\ z=\mu\end{array}(\lambda, \mu \in \mathbb{R})\right.\right. \) b) \( \left\{\begin{array}{l}x=3+2 \alpha \\ y=1-\alpha \\ z=5\end{array}\left\{\begin{array}{l}x=-1-6 \beta \\ y=3+3 \beta \\ z=5\end{array} \quad(\alpha, \beta \in \mathbb{R})\right.\right. \)