Circle And System Of Circles Question 269
Question: The radical centre of the circles $ x^{2}+y^{2}-16x+60=0,,x^{2}+y^{2}-12x+27=0, $ $ x^{2}+y^{2}-12y+8=0 $ is
[RPET 2000]
Options:
A) (13, 33/4)
B) (33/4, -13)
C) (33/4, 13)
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
$ S_1 $
$ \equiv $
$ x^{2}+y^{2}-16x+60=0 $ …..(i) $ S_2 $
$ \equiv $
$ x^{2}+y^{2}-12x+27=0 $ ….. (ii) $ S_3 $
$ \equiv $
$ x^{2}+y^{2}-12y+8=0 $ ….. (iii)
The radical axis of circle (i) and circle (ii) is $ S_1-S_2=0,\Rightarrow ,-4x+33=0 $ …. (iv) the radical axis of circle (ii) and circle (iii) is $ S_2-S_3=0 $
$ \Rightarrow -12+12y+19=0 $ …..(v)
Solving (iv) and (v), we get the radical centre $ ( \frac{33}{4},,\frac{20}{3} ) $ .
BETA
