Circle And System Of Circles Question 272
Question: The two circles $ x^{2}+y^{2}-2x+6y+6=0 $ and $ x^{2}+y^{2}-5x+6y+15=0 $
[Karnataka CET 2001]
Options:
A) Intersect
B) Are concentric
C) Touch internally
D) Touch externally
Show Answer
Answer:
Correct Answer: C
Solution:
Given, equations of the circles $ x^{2}+y^{2}-2x+6y+6 $ =0 …..(i) and $ x^{2}+y^{2}-5x+6y+15=0 $ …..(ii)
We know that the standard equation of a circle is $ x^{2}+y^{2}+2gx+2fy+c=0. $
Therefore for circle (i), $ g=-1;,f=3;,c=6; $ centre $ A=(1,,-3) $ and radius $ (r_1)=\sqrt{g^{2}+f^{2}-c}=\sqrt{1+9-6}=2 $ .
Similarly, for circle (ii), $ g=\frac{-5}{2};f=3;,c=15; $
Centre $ B\equiv ,( +\frac{5}{2},-3 ) $ and radius $ (r_2)=\sqrt{\frac{25}{4}+9-15}=\frac{1}{2} $
Therefore distance between A and B $ =\sqrt{{{( \frac{5}{2}-1 )}^{2}}+{{(-3+3)}^{2}}}=\frac{3}{2} $ and difference of radii $ (r_1-r_2)=2-\frac{1}{2}=\frac{3}{2}. $
Since distance between A and B is equal to $ r_1-r_2, $ therefore the circles touch each other internally.
BETA
