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

Asked by Abhisek | 1 year ago |  74

##### Solution :-

Centre ($$\dfrac{1}{2}$$, $$\dfrac{1}{4}$$) and radius $$\dfrac{1}{12}$$

Let us consider the equation of a circle with centre (h, k) and

Radius r is given as (x – h)+ (y – k)= r2

So, centre (h, k) = ($$\dfrac{1}{2}$$, $$\dfrac{1}{4}$$) and radius (r) =$$\dfrac{1}{12}$$

The equation of the circle is

(x – $$\dfrac{1}{2}$$)2 + (y – $$\dfrac{1}{4}$$)2 = ($$\dfrac{1}{12}$$)2

x2 – x + $$\dfrac{1}{4}$$ + y2 – $$\dfrac{y}{2}$$$$\dfrac{1}{16}$$$$\dfrac{1}{144}$$

x2 – x + $$\dfrac{1}{4}$$ + y2 – $$\dfrac{y}{2}$$ + $$\dfrac{1}{16}$$ = $$\dfrac{1}{144}$$

144x2 – 144x + 36 + 144y2 – 72y + 9 – 1 = 0

144x2 – 144x + 144y2 – 72y + 44 = 0

36x2 + 36x + 36y2 – 18y + 11 = 0

36x2 + 36y2 – 36x – 18y + 11= 0

The equation of the circle is 36x2 + 36y2 – 36x – 18y + 11= 0

Answered by Pragya Singh | 1 year ago

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