Question bank/JEE Main/Mathematics
JEE MainMathematics
The radius of the circle whose two normals are represented by the
equation
x2 - 5xy - 5x + 25y = 0 and which touches externally the circle x2 +
y2 - 2x + 4y - 4 = 0 is equal to
Show answer and solution
Normals are x2 - 5xy - 5x + 25y = 0 which can be factorised as,
(x - 5)(x - 5y) = 0
Hence x = 5 and y = 1 ⇒ centre of the circle is (5, 1)
and circle x2 + y2 - 2x + 4y - 4 = 0 is
(x - 1)2 + (y + 2)2 = 32
centre (1, - 2) and radius 3
external touch C1C2 = r1 + r2
5 = 3 + λ
so λ = 2
(x - 5)(x - 5y) = 0
Hence x = 5 and y = 1 ⇒ centre of the circle is (5, 1)
and circle x2 + y2 - 2x + 4y - 4 = 0 is
(x - 1)2 + (y + 2)2 = 32
centre (1, - 2) and radius 3
external touch C1C2 = r1 + r2
5 = 3 + λ
so λ = 2
Source: Grass-G / M-01 Mathematics Paper-1.docx
Need a video for this idea?
Scan any doubt from your notes and get a whiteboard explanation.