Equation of the circle which is such that the lengths of the tangents to it from the points (1,0), (0, 2) and
(3, 2) are 1, √7 and √2 respectively is
Answers
Equation of the circle which is such that the lengths of the tangents to it from the points (1,0), (0, 2) and (3, 2) are 1, √7 and √2 respectively is
(x - 7/3)² + (y - 5/12)² = (√137 / 12)²
Step-by-step explanation:
Let say Equation of circle
(x - h)² + (y - k)² = r²
h & k are center of circle & r = radius
lengths of the tangents to Circle from the points (1,0), (0, 2) and
(3, 2) are 1, √7 and √2 respectively
( h - 1)² + (k - 0)² = r² + 1²
=> ( h - 1)² + k² = r² + 1 Eq 1
( h - 0)² + (k - 2)² = r² + √7²
=> h² + (k-2)² = r² + 7 Eq 2
( h - 3)² + (k - 2)² = r² + √2²
=> (h-3)² + (k-2)² = r² + 2 Eq 3
Eq3 - Eq 2
=> (h-3)² - h² = -5
=> 9 - 6h = - 5
=> 6h = 14
=> 3h = 7
=> h = 7/3
Eq2 - Eq 1
h² - ( h - 1)² + (k-2)² - k² = 6
=> 2h - 1 + 4 - 4k = 6
=> 2(7/3) - 3 = 4k
=> 14 - 9 = 12k
=> k = 5/12
putting h & k in Eq 1
( h - 1)² + k² = r² + 1
( 7/3 - 1)² + (5/12)² = r² + 1
=> 16/9 + 25/144 = r² + 1
=> 256 + 25 = 144r² + 144
=> 144r² = 137
=> r = √137 / 12
(x - h)² + (y - k)² = r²
=> (x - 7/3)² + (y - 5/12)² = (√137 / 12)²