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What are the tangent equations to the circle x2 plus y2 -6x plus 4y plus 5 equals 0 at the points where it meets the x axis?
Wiki User
∙ 7y ago
x² + y² - 6x + 4y + 5 = 0
→ x² - 6x + 9 - 9 + y² + 4y + 4 - 4 + 5 = 0
→ (x - 3)² + (y + 2)² = 8
→ The circle has centre (3, -2)
It meets the x-axis when y = 0
→ x² + 0² - 6x + 4×0 + 5 = 0
→ x² - 6x + 5 = 0
→ (x - 1)(x - 5) = 0
→ x = 1 or 5.
- For x = 1:
→ slope of the tangent m' = -1/m = -1/-1 = 1
→ equation of tangent is:
y - 0 = 1(x - 1)
→ y = x - 1
- For x = 5:
→ slope of the tangent m' = -1/m = -1/1 = -1
→ equation of tangent is:
y - 0 = -1(x - 5)
→ y = -x + 5
→ y + x = 5
The tangents have equations y = x - 1 at (1, 0) and y + x = 5 at (5, 0)
Wiki User
∙ 7y ago
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1
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Does a secant tangent angle pass through the center of the circle?
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What are the tangent equations to the circle x2 plus y2 -6x plus 4y plus 5 equals 0 at the points where they meet the x axis?
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