Circle And System Of Circles Question 203
Question: The centre of a circle is (2, -3) and the circumference is $ 10\pi $ . Then the equation of the circle is
[Kerala (Engg.) 2002]
Options:
A) $ x^{2}+y^{2}+4x+6y+12=0 $
B) $ x^{2}+y^{2}-4x+6y+12=0 $
C) $ x^{2}+y^{2}-4x+6y-12=0 $
D) $ x^{2}+y^{2}-4x-6y-12=0 $
Show Answer
Answer:
Correct Answer: C
Solution:
Centre (2, - 3) Circumference $ =10\pi , $
therefore $ 2\pi r=10\pi $
therefore $ r=5. $ From $ {{(x-h)}^{2}}+{{(y-k)}^{2}}=r^{2} $ , $ {{(x-2)}^{2}}+{{(y+3)}^{2}}=5^{2} $
therefore $ x^{2}+y^{2}-4x+6y+13=25 $
therefore $ x^{2}+y^{2}-4x+6y-12=0 $ , which is the required equation of the circle.
BETA
