Answer:
b) on the circle
The point lies on the circle S =0
Step-by-step explanation:
Step(i):-
Given center of the circle = H(4,0)
Radius of the circle 'r' = 10
Equation of the circle
[tex](x -h)^{2} +(y-k)^{2} = r^{2}[/tex]
[tex](x -4)^{2} +(y-0)^{2} = 10^{2}[/tex]
[tex](x )^{2} -8 x + 16 + (y)^{2} = 10^{2}[/tex]
[tex](x )^{2} -8 x + (y)^{2} = 10^{2} - 16[/tex]
[tex]x ^{2} -8 x + y^{2} = 84[/tex]
Step(ii):-
S =0 is a circle and P(x₁ , y₁) be a point in its plane
Then (i) P lies inside the circle S = 0 ⇔ S₁₁ < 0
ii) P lies outside the circle S = 0 ⇔ S₁₁ > 0
iii) P lies on the circle S = 0 ⇔ S₁₁ = 0
now
[tex]S_{11} = x^{2} _{1} + y^{2} _{1} + 2 g x_{1} + 2 f y_{1} +c[/tex]
[tex]S_{11} = x_{1} ^{2} -8 x_{1} + y_{1} ^{2} = 84[/tex]
Given point ( -2 , 8)
[tex]S_{11} = (-2) ^{2} -8 (-2) + (8) ^{2} -84 = 0[/tex]
P lies on the circle S = 0 ⇔ S₁₁ = 0
