Answer:
(-4,0) and (4,0)
Option C and Option F are the right options.
solution,
[tex] {x}^{2} + {y}^{2} = 16......(equation \: i) \\ \frac{ {x}^{2} }{ {4}^{2} } - \frac{ {y}^{2} }{ {4}^{2} } = 16......( \: equation \: ii) \\ from \: equation \: ii \\ \frac{ {x}^{2} }{ {4}^{2} } - \frac{ {y}^{2} }{ {4}^{2} } = 1 \\ \frac{ {x}^{2} - {y}^{2} }{ {4}^{2} } = 1 \\ \frac{ {x}^{2} - {y}^{2} }{16} = 1 \\ {x}^{2} - {y}^{2} = 16......... \: equation \: iii \\ now \: adding \: equation \: i \: and \: ii \\ {x}^{2} + {y}^{2} = 16 \\ {x}^{2} - {y}^{2} = 16 \\ ...................... \\ 2 {x}^{2} = 32 \\ or \: {x}^{2} = \frac{32}{2} \\ or \: {x}^{2} = 16 \\ or \: x = \sqrt{16} \\ or \: x = \sqrt{ {(4)}^{2} } \: \: \: or \: \: \sqrt{ {( - 4)}^{2} } \\ x = + 4 \: or \: - 4[/tex]
Now
When X=4,
[tex] {y}^{2} = 16 - {x}^{2} \\ {y}^{2} = 16 - {4}^{2} \\ {y}^{2} = 16 - 16 \\ {y = 0}^{2} \\ y = \sqrt{ {(0)}^{2} } \\ y = 0[/tex]
When X=-4,
[tex] {y}^{2} = 16 - {x}^{2} \\ {y}^{2} = 16 - {( - 4)}^{2} \\ {y}^{2} = 16 - 16 \\ {y}^{2} = 0 \\ y = \sqrt{ {(0)}^{2} } \\ y = 0[/tex]
The solutions are (4,0) and (-4,0)
Hope this helps...
good luck on your assignment..
