JEE Advanced | 2014CircleHard
Question
A circle S passes through the point (0, 1) and is orthogonal to the circles (x - 1)2 + y2 = 16 and x2 + y2 = 1. Then
Options
A.Radius of S is 8
B.Radius of S is 7
C.Centre of S is (-7, 1)
D.Centre of S is (-8, 1)
Solution
Let the circle be x2 + y2 + 2gx + 2fy + C = 0.
It is orthogonal with (x - 1)2 + y2 = 16
∴ 2(-g + 0) = - 15 + C ⇒ - 2g = - 15 + C
It is also orthogonal with x2 + y2 = 1
∴ 0 = - 1 + C ⇒ C = 1
∴ g = 7
This circle passes through (0, 1)
∴ 1 + 2f + 1 = 0 ⇒ f = - 1
∴ The circle is x2 + y2 + 14x - 2y + 1 = 0
(x + 7)2 + (y - 1)2 = 49
∴ centre : (-7, 1) and radius : 7
It is orthogonal with (x - 1)2 + y2 = 16
∴ 2(-g + 0) = - 15 + C ⇒ - 2g = - 15 + C
It is also orthogonal with x2 + y2 = 1
∴ 0 = - 1 + C ⇒ C = 1
∴ g = 7
This circle passes through (0, 1)
∴ 1 + 2f + 1 = 0 ⇒ f = - 1
∴ The circle is x2 + y2 + 14x - 2y + 1 = 0
(x + 7)2 + (y - 1)2 = 49
∴ centre : (-7, 1) and radius : 7
Create a free account to view solution
View Solution FreeMore Circle Questions
The tangent from the point of intersection of the lines 2x - 3y + 1 = 0 and 3x - 2y - 1 = 0 to the circle x2 + y2 + 2x -...Two lines through (2, 3) from which the circle x2 + y2 = 25 intercepts chords of length 8 units have equations...For the given circles x2 + y2 − 6x − 2y + 1 = 0 and x2 + y2 + 2x − 8y + 13 = 0, which of the following...Two circles x2 + y2 = 6 and x2 + y2 - 6x + 8 = 0 Then the equation of the circle through their points of intersection an...The length of the tangent drawn from the point (2, 3) to the circles 2(x2 + y2) - 7x + 9y -11 = 0 -...