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3.  Find the equation of the circle with

   centre \left(\frac{1}{2},\frac{1}{4} \right ) and radius \frac{1}{12}

 

Answers (1)

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As we know, 

The equation of the circle with centre ( h, k) and radius r is given by:

(x-h)^2+(y-k)^2=r^2

So, given here

(h,k)=\left ( \frac{1}{2},\frac{1}{4} \right ) 

AND 

 r=\frac{1}{12}

So, the equation of a circle is:

\left ( x-\frac{1}{2}\right )^2+\left ( y-\frac{1}{4}\right )^2=\left ( \frac{1}{12}\right )^2

x^2-x+\frac{1}{4}+y^2-\frac{1}{2}y+\frac{1}{16}=\frac{1}{144}

x^2+y^2-x-\frac{1}{2}y-\frac{11}{36}=0

36x^2+36y^2-36x-18y-11=0

Posted by

Pankaj Sanodiya

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