logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XI  >>  Math  >>  Conic Sections
0 votes

Find the equation of the circle with centre $(-2,3)$ and radius $4$.

$\begin{array}{1 1} x^2+y^2+4x-6y-3=0 \\ x^2+y^2+4x+6y-3=0 \\ x^2+y^2+4x-6y+3=0\\ x^2+y^2-4x-6y-3=0\end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • Given C (h, k) be the centre and r the radius of circle. Let P(x, y) be any point on the circle. Then, by the definition, | CP | = r . By the distance formula, we have, $(x-h)^2+(y-k)^2=r^2$
  • http://clay6.com/mpaimg/Toolbar_1.png
Given: circle with centre (-2,3) and radius 4, $\rightarrow r = 4, h = -2, k = 3$
Therefore equation of circle using the distance formula is: $ (x-(-2))^2 + (y-3)^2 = 4^2$
$\quad \quad x^2+4x+4+y^2-6y+9=16$
$\quad \quad x^2+y^2+4x-6y-3=0$
answered Apr 1, 2014 by balaji.thirumalai
edited Apr 1, 2014 by balaji.thirumalai
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...