Using the formula given below, we find length of tangent drawn from the point (x1, y1).
Length of the tangent = √(x12+y12+2gx1+2fy1+c)
Note :
(i) If the length is 0, then we say the given point must be on the circle.
(ii) If the length is > 0, then we say the point must be outside the circle.
(iii) If the length is < 0, then we say the point must be inside the circle.
Example 1 :
Find the length of tangent to the circle
x2+y2-4x-3y+12 = 0
from the point (2,3)
Solution :
Length of the tangent = √(x12+y12+2gx1+2fy1+c)
Here x1 = 2 and y1 = 3
= √22+32-4(2)-3(3)+12
= √(4+9-8-9+12)
= √(4+12-8)
= √8
= 2√2 units
Example 2 :
Show that the point (2,-1) lies out side the circle
x2+y2-6x-8y+12 = 0
Solution :
Here x1 = 2 and y1 = -1
= √22+(-1)2-6(2)-8(-1)+12
= √(4+1-16+8+12)
= √(5+8+12+1-8)
= √(26-8)
= √18
= 3√2 units
The length of tangent is positive. So, the given point lies outside of the circle.
Example 3 :
Find the length of tangent to the circle
x2+y2-4x+8y-5 = 0
from the point (2, 1)
Solution :
Here x1 = 2 and y1 = 1
= √22+12-4(2)+8(1)-5
= √(4+1-8+8-5)
= 0
The length of tangent is zero. So, the given point lies on the circle.
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