Circle And System Of Circles Question 277
Question: If the two circles $ 2x^{2}+2y^{2}-3x+6y+k=0 $ and $ x^{2}+y^{2}-4x+10y+16=0 $ cut orthogonally, then the value of k is
[Kerala (Engg.) 2002]
Options:
A) 41
B) 14
C) 4
D) 0
Show Answer
Answer:
Correct Answer: C
Solution:
Given circles are $ 2x^{2}+2y^{2}-3x+6y+k=0 $ or $ x^{2}+y^{2}-\frac{3}{2}x+3y+\frac{k}{2}=0 $ …..(i) and $ x^{2}+y^{2}-4x+10y+16=0 $ …..(ii)
Circle (i) and (ii) cut orthogonally, then $ 2g_1,g_2+2f_1f_2=c_1+c_2 $
$ 2( -\frac{3}{4} ),(-2)+2( \frac{3}{2} ),.,5=\frac{k}{2}+16 $
$ 3+15=\frac{k}{2}+16 $
therefore $ 18=\frac{k}{2}+16 $
therefore $ k=4. $
BETA
