Respuesta :
Answer:
Point [tex]\left(1,\sqrt{3}\right)[/tex] lies inside the circle.
Step-by-step explanation:
To determine the point which lies outside the circle, calculate the distance between the points and compare that value with radius of circle.
If d<r point lies inside the circle, d>r point lies outside circle and d=r point is on the circle.
Let O be the center of origin of the circle. So, [tex]O=\left ( 0,0 \right )[/tex].
Also given that circle contain point [tex]\left ( 3,0 \right )[/tex]. So, [tex]x=3,y=0[/tex]
Now equation of circle having center (h,k) is given by the equation,
[tex]\left(x-h\right)^2+\left(y-k\right)^2=r^2[/tex]
[tex]h=0,k=0[/tex]
Substituting the value,
[tex]\left(3-0\right)^2+\left(0-0\right)^2=r^2[/tex]
[tex]9=r^2[/tex]
[tex]r=\pm 3[/tex]
Now calculate the distance between point [tex]\left(3,0\right)[/tex] and [tex]\left (1,\sqrt{3}\right)[/tex].
Distance formula between points [tex]\left(x_{1},y_{1}\right)[/tex] and [tex]\left(x_{2},y_{2}\right)[/tex]is given as,
[tex]d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}[/tex]
Now, [tex]x_{1}=1,y_{1}=\sqrt{3},x_{2}=3,y_{2}=0[/tex]
[tex]\therefore d=\sqrt{\left(3-1\right)^2+\left(0-\sqrt{3}\right)^2}[/tex]
Simplifying,
[tex]\therefore d=\sqrt{\left(2\right)^2+\left(-\sqrt{3}\right)^2}[/tex]
[tex]\therefore d=\sqrt{4+3}[/tex]
[tex]\therefore d=\sqrt{7}[/tex]
[tex]\therefore d=2.64[/tex]
Since r = 3 and d = 2.64. That is, d < r.
So point [tex]\left (1,\sqrt{3}\right)[/tex] lies inside the circle.
