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Find the equation of the circle with centre ($\large\frac{1}{2}$, $\large\frac{1}{4}$) and radius $\large\frac{1}{12}$

$\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} $

1 Answer

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 (0,2) and radius 2, $\rightarrow r = 4, h = -2, k = 3$
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
 
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