the inverse point of (2,-3) w.r.t circle x*2+y*2+6x-4y-12=0 is
Answers
Given : Circle x² + y² + 6x - 4y - 12 = 0 and a point (2,-3)
To find : Inverse point of ( 2, -3) w.r.t circle
Solution:
x² + y² + 6x - 4y - 12 = 0
=> (x + 3)² - 9 + (y - 2)² - 4 - 12 = 0
=> (x + 3)² + (y - 2)² = 25
=> (x + 3)² + (y - 2)² = 5²
=> Center = - 3 , 2
Radius = 5
O = (- 3 , 2)
P = ( 2, - 3)
OP = √(-3 - 2)² + (2 - (-3))²) = 5√2
P' is inverse of point P
P' is point on OP such that
| OP | | OP'| = radius²
=> | 5√2| | OP'| = 25
=> | OP'| = 5/√2
OP Slope = - 1
y - 2 = - 1(x -(-3))
=> y - 2 = - x - 3
=> x + y = - 1
Let say P' = h , k
h + k = -1 => h = - 1 - k
O = (-3 , 2)
|OP'| = √(h - (-3))² + (k - 2)²
= √(-1 - k + 3))² + (k - 2)²
=> √(2 - k)² + (k - 2)²
=> √2 | k - 2 |
√2 | k - 2 | = 5/√2
=> | k - 2 | = 5/2
=> k = 9/2 , - 1/2
k lies between 2 , - 3
hence k = - 1/2
h = - (1 + k)
=> h = - (1 - 1/2)
=> h = -1/2
(h , k) = (-1/2 , -1/2)
(-1/2 , -1/2) is the inverse point of (2,-3) w.r.t circle x² + y² + 6x - 4y - 12 = 0
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