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Determine whether the given point is interior, exterior, or on the circle [tex][tex]$(x+22)^2+(y-3)^2=81$[/tex][/tex].

Point: [tex]Q(3,0)[/tex]


Sagot :

To determine the position of the point [tex]\( Q(3,0) \)[/tex] relative to the circle given by the equation [tex]\((x+22)^2 + (y-3)^2 = 81\)[/tex], we need to follow these steps:

1. Identify the center and radius of the circle:

The general form of a circle equation is [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex], where [tex]\((h, k)\)[/tex] is the center and [tex]\(r\)[/tex] is the radius.

From the given equation [tex]\((x + 22)^2 + (y - 3)^2 = 81\)[/tex], we observe that:
- The center [tex]\((h, k)\)[/tex] is [tex]\((-22, 3)\)[/tex].
- The right-hand side of the equation represents [tex]\(r^2\)[/tex], which is 81. Thus, the radius [tex]\(r\)[/tex] is [tex]\(\sqrt{81} = 9\)[/tex].

2. Calculate the distance between the point [tex]\( Q(3, 0) \)[/tex] and the center of the circle [tex]\((-22, 3)\)[/tex]:

The distance formula between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:
[tex]\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]

Applying the coordinates:
[tex]\[ \text{Distance} = \sqrt{(3 - (-22))^2 + (0 - 3)^2} \][/tex]
Simplifying inside the square root:
[tex]\[ \text{Distance} = \sqrt{(3 + 22)^2 + (0 - 3)^2} \][/tex]
[tex]\[ \text{Distance} = \sqrt{25^2 + (-3)^2} \][/tex]
[tex]\[ \text{Distance} = \sqrt{625 + 9} \][/tex]
[tex]\[ \text{Distance} = \sqrt{634} \][/tex]
Approximating the square root:
[tex]\[ \text{Distance} \approx 25.179 \][/tex]

3. Compare this distance to the radius of the circle:

- The radius of the circle is 9.
- The calculated distance from the point [tex]\( Q \)[/tex] to the center of the circle is approximately 25.179.

Since the distance (25.179) is greater than the radius (9), the point [tex]\( Q(3,0) \)[/tex] lies outside the circle.

4. Conclusion:

The given point [tex]\( Q(3,0) \)[/tex] is in the exterior of the circle [tex]\((x+22)^2 + (y-3)^2 = 81\)[/tex].