Compute the derivative dy/dx and check where it is zero (for horizontal tangents) or undefined (for vertical tangents).
2x² + 2y² - 20x + 12y + 3 = 0
d/dx [2x² + 2y² - 20x + 12y + 3] = 0
4x + 4y dy/dx - 20 + 12 dy/dx = 0
(4y + 12) dy/dx = 20 - 4x
(y + 3) dy/dx = 5 - x
dy/dx = (5 - x) / (y + 3)
• Horizontal tangents:
dy/dx = 0 → 5 - x = 0 → x = 5
Solve for y when x = 5 :
2•5² + 2y² - 20•5 + 12y + 3 = 0
2y² + 12y - 47 = 0
y = (-6 ± √(130))/2
So there are two horizontal tangents at the points
(5, (-6 - √(130))/2) and (5, (-6 + √(130))/2)
• Vertical tangents:
1/(dy/dx) = 0 → y + 3 = 0 → y = -3
Solve for x when y = -3 :
2x² + 2•(-3)² - 20x + 12•(-3) + 3 = 0
2x² - 20x - 15 = 0
x = (10 ± √(130))/2
So there are two vertical tangents at the points
((10 - √(130))/2, -3) and ((10 + √(130))/2, -3)
Alternatively, you can complete the square to identify the equation of a circle:
2x² + 2y² - 20x + 12y + 3 = 0
2 (x² - 10x) + 2 (y² + 6y) = -3
2 (x² - 10x + 25 - 25) + 2 (y² + 6y + 9 - 9) = -3
2 (x - 5)² - 50 + 2 (y + 3)² - 18 = -3
2 (x - 5)² + 2 (y + 3)² = 65
(x - 5)² + (y + 3)² = 65/2
which is a circle centered at (5, -3) with radius √(65/2). The horizontal tangents occur at the points where the x term vanishes (x = 5), and the vertical ones where y vanishes (y = -3).