Math, asked by soumithnanikumar1, 9 months ago

For the circle ax^2+y^2+bx+dy+2=0 centre is (1,2) then 2b+3d=...

Answers

Answered by amitnrw
11

Given :  Circle x² + y² + bx + dy + 2 = 0  , centre is (1,2)

To Find : 2b+3d=

Solution:

x² + y² + bx + dy + 2 = 0

=> x² + bx   + y² + dy  + 2 = 0

=> x²  + bx  + (b/2)² - (b/2)²   +  y² + dy  +   (d/2)² - (d/2)²  + 2 = 0

=> (x + b/2)² + (y + d/2)²  =   (b/2)² +  (d/2)² - 2

=> (x - (-b/2))² +  (y - (-d/2))² = ( b² + d² - 8)/4

Comparing with

(x - h)² + (y - k)² = r²

h , k   center of circle

=>  h = -b/2  = 1  => b = -2

k = -d/2 = 2  => d = - 4

2b + 3d  =   2 * (-2) + 3(- 4)

= - 4 - 12

= - 16

( b² + d² - 8)/4 = r²

=>  r² = 3

=> radius = √3

2b + 3d  =  - 16

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