Question bank/JEE Main/Mathematics
JEE MainMathematics
If the locus of the centre of the variable circle which intersects
the circles x2 + y2 = 4 and
x2 + y2 + 2x + 4y = 6 orthogonally has the equation ax + by = 1, then
the value of [IMAGE] is equal to
Show answer and solution
Let the required circle be x2 + y2 + 2gx + 2fy + c = 0
It is orthogonal with x2 + y2 = 4
∴ c - 4 = 0
∴ c = 4
and also with x2 + y2 + 2x + 4y - 6 = 0
∴ 2(1 · g + 2f) = 4 - 6 ⇒ 2(g + 2f) = - 2 ⇒ g + 2f = - 1
∴ locus of the centre is
- x - 2y = - 1 ⇒ x + 2y - 1 = 0
⇒ a = 1 and b = 2
Now [IMAGE] On putting [IMAGE] = t, we get [IMAGE] Put [IMAGE] = - ln 2
It is orthogonal with x2 + y2 = 4
∴ c - 4 = 0
∴ c = 4
and also with x2 + y2 + 2x + 4y - 6 = 0
∴ 2(1 · g + 2f) = 4 - 6 ⇒ 2(g + 2f) = - 2 ⇒ g + 2f = - 1
∴ locus of the centre is
- x - 2y = - 1 ⇒ x + 2y - 1 = 0
⇒ a = 1 and b = 2
Now [IMAGE] On putting [IMAGE] = t, we get [IMAGE] Put [IMAGE] = - ln 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.