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Find the intersection of the circle x^2+ y^2 = 29 and y = x - 3 algebraically

User Pixeladed
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2 Answers

13 votes
13 votes

Answer:

Intersections: (-2, -5) and (5, 2)

Hope this helps!

Find the intersection of the circle x^2+ y^2 = 29 and y = x - 3 algebraically-example-1
User Umeboshi
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25 votes

Final answer:

To find the intersections, substitute y from the line equation y = x - 3 into the circle equation x^2 + y^2 = 29 and solve for x. Then, use the found x-values to solve for the corresponding y-values. The points of intersection are (-2, -5) and (5, 2).

Step-by-step explanation:

To find the intersection of the circle x^2 + y^2 = 29 and the line y = x - 3, we substitute the expression for y from the line equation into the circle equation.

Substitute y = x - 3 into x^2 + y^2 = 29:
x^2 + (x - 3)^2 = 29.

Expand the squared term and simplify:
x^2 + x^2 - 6x + 9 = 29.

Combine like terms and solve for x:
2x^2 - 6x - 20 = 0.

Factor the quadratic equation or use the quadratic formula to find the x-values.
(Let's assume it factored to (2x+4)(x-5) = 0, then x can be -2 or 5).

Substitute the x-values into y = x - 3 to find corresponding y-values:
For x = -2, y = -5 and
for x = 5, y = 2.

The points of intersection are (-2, -5) and (5, 2).

User Chase Henslee
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