For the circle ax^2+y^2+bx+dy+2=0 centre is (1,2) then 2b+3d=...
Answers
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
Learn more:
find the equation of circle which passed through origin and cuts off ...
https://brainly.in/question/2075668
Equation of the circle which is such that the lengths of the tangents ...
https://brainly.in/question/13068835
find the equation of a circle the endpoints of whose one diameter are ...
https://brainly.in/question/13194415