Question

plssss nohelp me plssss

answer well

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Answer 1

Answer:-

The ordered pair not lie on the circle

Further Explanation:

we know that

The equation of the circle in standard form is equal to

$(x-h)^2+(y-k)^2=r^2(x−h)$

where

(h,k) is the center

In this problem we have

(h,k)=(0,0)

substitute

$(x)^2+(y)^2=r^2(x)$

Find the radius

Remember that

The distance between the center and any point on the circle is equal to the radius

we have

(0,0) and (3,0)

The distance is equal to

$d=\sqrt{(3-0)^2+(0-0)^2}$

therefore

The radius is 3 units

The equation of the circle is equal to

(x)^2+(y)^2=3^2(x) 2 +(y) 2 =3x^2+y^2=9x

2 +y 2 =9

Determine if the point $(2,\sqrt{6})(2, 6 )$ lie on the circle

If a ordered pair lie on the circle, then the ordered pair must satisfy the equation of the circle

substitute the value of x and the value of y in the equation

$2^2+(\sqrt{6})^2=92$

10=910=9 => is not true

The ordered pair not satisfy the equation

Therefore ,the ordered pair not lie on the circle