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Find the points on the line x=1
where the circle with equation
2x²+2y²-5x+7y-36-0 intersect

User Draeron
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1 Answer

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Answer: To find the points where the circle intersects the line x=1, we substitute x=1 in the equation of the circle:

2(1)² + 2y² - 5(1) + 7y - 36 = 0

Simplifying, we get:

2y² + 7y - 31 = 0

We can solve this quadratic equation by using the quadratic formula:

y = (-7 ± √(7² - 4(2)(-31))) / (2(2))

y = (-7 ± √225) / 4

y = (-7 ± 15) / 4

So the two possible values of y are:

y = 2 or y = -8/2 = -4

Therefore, the points where the circle intersects the line x=1 are (1, 2) and (1, -4).

Explanation:

User Sumshyftw
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