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At every point (x,y) on a curve, the normal line passes through the fixed point (3,0). If the curve also goes through (3,4), then the equation of the curve is
- x² + y² + 6x − 7 = 0
- 2(x² + y²) − 12x − 7 = 0
- x² + y² − 6x − 7 = 0
- None of these
Correct answer: x² + y² − 6x − 7 = 0
Solution
The correct option describes a circle centered at (3,0) with a radius that allows it to pass through the point (3,4). The equation can be rearranged to show that it satisfies the condition of having the normal line at any point on the curve intersect the fixed point.
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