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4 years ago

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(1, 1/2, 1/3, 1/4, ...)

1 answer:

SpyIntel [72]4 years ago

4 0

(1, 1,2 1,3 1,4 1,5 1,6)

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3 years ago

Does the point (1,√3) lie inside outside or on the circle with a center at the origin containing the point (3,0)? Question 2 opt

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Answer:

Point $\left(1,\sqrt{3}\right)$ 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, $O=\left ( 0,0 \right )$ .

Also given that circle contain point $\left ( 3,0 \right )$ . So, $x=3,y=0$

Now equation of circle having center (h,k) is given by the equation,

$\left(x-h\right)^2+\left(y-k\right)^2=r^2$

$h=0,k=0$

Substituting the value,

$\left(3-0\right)^2+\left(0-0\right)^2=r^2$

$9=r^2$

$r=\pm 3$

Now calculate the distance between point $\left(3,0\right)$ and $\left (1,\sqrt{3}\right)$ .

Distance formula between points $\left(x_{1},y_{1}\right)$ and $\left(x_{2},y_{2}\right)$ is given as,

$d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

Now, $x_{1}=1,y_{1}=\sqrt{3},x_{2}=3,y_{2}=0$

$\therefore d=\sqrt{\left(3-1\right)^2+\left(0-\sqrt{3}\right)^2}$

Simplifying,

$\therefore d=\sqrt{\left(2\right)^2+\left(-\sqrt{3}\right)^2}$

$\therefore d=\sqrt{4+3}$

$\therefore d=\sqrt{7}$

$\therefore d=2.64$

Since r = 3 and d = 2.64. That is, d < r.

So point $\left (1,\sqrt{3}\right)$ lies inside the circle.

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