cyrusM3SO
contestada

ng
10
8
6
4
2
10 8 64-2
-2
4
6
-8
618
-10
ctrich ar Of EICS
y
2 4 6 8 10
x
Ta OnciC
A circle centered at the origin contains the point
(0, -9). Does (8, √17) also lie on the circle? Explain.
O No, the distance from the center to the point
(8, √17) is not the same as the radius.
O No, the radius of 10 units is different from the
distance from the center to the point
(8, √17).
O Yes, the distance from the origin to the point
(8, √17) is 9 units.
O Yes, the distance from the point (0, -9) to the point
(8, √17) is 9 units.

ng 10 8 6 4 2 10 8 642 2 4 6 8 618 10 ctrich ar Of EICS y 2 4 6 8 10 x Ta OnciC A circle centered at the origin contains the point 0 9 Does 8 17 also lie on the class=

Respuesta :

Approximate value of √17

  • ~4+

So the point is (8,4)

As we can see in graph the point is contained by the circle

Hence the distance

  • √(0-8)²+(9-4)²
  • √8²+5²
  • √89

Approximately 9 units

Option D

semsee45

Answer:

Yes, the distance from the origin to the point (8, √17) is 9 units

Step-by-step explanation:

Equation of a circle

[tex](x-a)^2+(y-b)^2=r^2[/tex]

where:

  • (a, b) is the center
  • r is the radius

Given:

  • center = (0, 0)
  • r = 9

Substitute the given values into the formula to create the equation of the circle:

[tex]\implies (x-0)^2+(y-0)^2=9^2[/tex]

[tex]\implies x^2+y^2=81[/tex]

To find if point [tex]\sf (8, \sqrt{17})[/tex] lies on the circle, substitute x = 8 into the found equation and solve for y:

[tex]\implies 8^2+y^2=81[/tex]

[tex]\implies 64+y^2=81[/tex]

[tex]\implies y^2=81-64[/tex]

[tex]\implies y^2=17[/tex]

[tex]\implies y=\sqrt{17}[/tex]

As y = √17  when x = 8, this proves that point (8, √17) lies on the circle.

The distance from the origin to any point on the circle is the radius.

Therefore, as the radius is 9 units, the distance from the origin to the point (8, √17) is 9 units.

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