A circle S=0 with radius √2 touches the line
x+y-2=0 at (1,1). Then the length of the tangent drawn
from the point (1, 2) to S=0=?
Answers
Given : A circle S=0 with radius √2 touches the line x+y-2=0 at (1,1).
To Find : length of the tangent drawn from the point (1, 2) to S=0
Solution:
x + y - 2 = 0
touches at point ( 1 , 1)
Normal to x + y - 2 = 0 will have slope = 1
y - 1 = 1(x - 1)
=> y = x is the normal to x + y - 2 = 0 ( tangent ) Hence passes through center
radius = √2
(x - 1)² + ( y - 1)² = ( √2)²
y = x
=> (x - 1)² + (x - 1)² = ( √2)²
=> (x - 1)² = 1
=> x = 0 , 2
=> y = 0 , 2
Two possible circle
x² + y² = 2 or ( x - 2)² + ( y - 2)² = 2
point ( 1 , 2)
=> 1² + 2² = 5 > 2 hence lies out side circle
( 1 - 2)² + (2 - 2)² = 1 < 2 hence lies inside the circle
so ( x - 2)² + ( y - 2)² = 2 is eliminated
Hence x² + y² = 2 is the circle
point = ( 1, 2)
x² + y² = 2
=> dy/dx = - x/y
Tangency point ( h , k) on circle
=> ( k - 2)/ (h - 1) = - h/k Equating slope
=> k² - 2k = -h² + h
=> k² + h² = 2k + h
h² + k² = 2
=> 2k + h = 2
=> h = 2 - 2k
(2 - 2k)² + k² = 2
=> 5k² - 8k + 2 = 0
=> k = (8 ± √64 - 40)/2(5)
=> k =( 4 ± √6 )/5
k = (4 + √6)/5 , (4 - √6)/5
h = 2 - 2k
h = (2 - 2√6)/5 , 2 + 2√6
points of tangency are :
( (2 - 2√6)/5 , (4 + √6)/5) and ( (2 + 2√6)/5 , (4 - √6)/5 )
Distance from ( 1, 2)
= √ {( (2 - 2√6)/5 - 1) ² + ((4 + √6)/5 - 2)² }
= √ {( (-3 - 2√6)/5 ) ² + ((-6 + √6)/5)² }
= (1/5) √ {( 9 + 24 + 12√6 + 36 + 6 - 12√6 }
= (1/5) √ {(75 }
= √3
the length of the tangent drawn from the point (1, 2) to S=0 is √3
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