Question
Question: The equation of the radical of a coaxial system of circle whose limiting points are (2, – 1) and (–3...
The equation of the radical of a coaxial system of circle whose limiting points are (2, – 1) and (–3, 2), is –
A
5x + 3y – 4 = 0
B
5x + 3y + 4 = 0
C
5x – 3y + 4 = 0
D
3x – 5y + 4 = 0
Answer
5x – 3y + 4 = 0
Explanation
Solution
The equations of circles having (2, –1) and
(–3, 2) as limiting points are
S1 ŗ (x – 2)2 + (y + 1)2 = 0
and S2 ŗ (x + 3)2 + (y – 2)2 = 0
Clearly, (1) and (2) are members of the coaxial system of circles whose limiting points are
(2, –1) and (–3, 2).
So, the equation of the radical axis is S2 – S1 = 0
Ž 10x – 6y + 8 = 0
Ž 5x – 3y + 4 = 0
Hence (3) is the correct answer.