Question

the inverse point of (2,-3) w.r.t circle x*2+y*2+6x-4y-12=0 is​

Answer 1

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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