IDNLearn.com: Your go-to resource for finding expert answers. Discover prompt and accurate answers from our experts, ensuring you get the information you need quickly.
Answer:
Answer is attached in images
Step-by-step explanation:
Given:
x+z=1 and
[tex]z=4-x^{2} -y^{2}[/tex]
Now,
[tex]1-x=4-x^{2} 2-y^{2} 2\\x^{2} -x+y^{2} \\x^{2} -x+1/4+y^{2} =3\\(x-1/2)^{2} +y^{2} =3+1/4\\(x-1/2)^{2} +y^{2} =(\sqrt(13)/2)^{2}[/tex]
Which is circle with center (1/2,0) radius
[tex]\sqrt{13/2}[/tex] take, [tex]x=1/2+\sqrt{3} /2 cos\alpha \\y=\sqrt{13}/2 sin\alpha 0\leq \alpha \leq 2\pi[/tex]
Now,
