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Lady bird [3.3K]
4 years ago
8

Given the circle with the equation (x - 3)^2 + y^2 = 49, determine the location of each point with respect to the graph of the c

ircle. In your final answer, state whether each point is on the interior, exterior, or circumference of the circle. Include your calculations as proof of each point’s location. PLEASE PUT THE CALCULATIONS
A. (-1, 1)
B. (10, 0)
C. (4, -8)
Mathematics
2 answers:
Fofino [41]4 years ago
7 0

Point A (-1.1)  Plug this into the equation:

(-1 - 3)^2 + 1^1  =  17  so it cant be on the circumference because it does not satisfy the equation.

The radius of the circle is sqrt49 = 7   and the center is at (3,0)  so the point A is inside the circle. The end points of the circle are at x = -4 and 10 and y endpoints are at y=7 and -7.

Point (10,0)  Plug this in:-

(10-3)^2 + 0^2 = 49

This satisfies the equation so its on the circumference.

Pointt  (4, -8)

The minimum point of the circle will be at  (3, -7)  because the center is at (3,0) and the radius is 7. so the point (4, -8) will be outside the circle.

Nadya [2.5K]4 years ago
5 0

The generic expression for a circle is

(x-x_0)^2 + (y-y_0)^2 = r^2

where (x_0,y_0) is the center, and r is the radius.

If you compare this expression to yours, you can see that you have

(x-3)^2 + (y-0)^2 = 7^2

So, the center of the circle is (3,0) and the radius is 7.

Now, to state whether a point P is inside, outside or in the circumference, you can compare the distance from the center C and the radius:

  • If PC > r, the point P is outside the circle
  • If PC < r, the point P is inside the circle
  • If PC = r, the point P is on the circumference of the circle

Remember that, if you want to compute the distance between two points A = (A_x, A_y),\ B = (B_x, B_y), you can use the formula

AB = \sqrt{(A_x-B_x)^2+(A_y-B_y)^2}

Using this formula, you can see that (if we call the center of the cirlce O):

  • AO = \sqrt{17} \approx 4.1 < 7, so this point is inside the circle
  • BO = 7, so this point is on the circumference
  • CO = \sqrt{65} = 8.06>7, so this point is outside the circle
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