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Help! About Conic Section (Circle)?
Given a circle with eq X^2 + Y^2 - 2X - 6Y + 6=0
How to prove that point (3,5) lies outside the circle?
b2-4ac>0?
1 Answer
- poornakumar bLv 78 years agoFavorite Answer
x² + y² - 2x - 6y + 6=0,
[(x -1)² -1]+[(y -3)² -9] -4 =0
in the standard form
(x -xₒ)² +(y -yₒ)² -r² =0.
It is a circle with radius 2, centre (1, 3).
For a point within the circle its distance to its center shouldn't exceed the radius.
In this instance the point is (3,5). Its distance from center (1, 3) is
d² = (3 -1)² +(5 -3)² = 8,
d = 2√2 > 2.
Hence the point is outside the circle.