Question

Type 4 points found on the circumference of the following equation. Exact points only, no approximations.

Answer 1

The points on the circumference are (3, 5), (3, -9), (8, -2 + √24) and (8, -2 - √24))

How to determine the points on the circumference of the circle?

The circle equation is given as:

(x-3)^2 +(y+2)^2= 49

Rewrite as:

(y+2)^2= 49 -(x-3)^2

Take the square root of both sides

y+2= ±√[49 -(x-3)^2]

Subtract 2 from both sides

y = -2 ± √[49 -(x-3)^2]

Next, we determine the points

Set x = 3

y = -2 ± √[49 -(3-3)^2]

Evaluate

y = -2 ± 7

Solve

y = 5 and y = -9

So, we have

(x, y) = (3, 5) and (3, -9)

Set x = 8

y = -2 ± √[49 -(8-3)^2]

Evaluate

y = -2 ± √24

Solve

y = -2 - √24 and y = -2 + √24

So, we have

(x, y) = (8, -2 + √24) and (8, -2 - √24)

Hence, the points on the circumference are (3, 5), (3, -9), (8, -2 + √24) and (8, -2 - √24)

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brainly.com/question/1559324

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