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

Asked by Aaryan | 1 year ago |  79

##### Solution :-

Given:

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

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 Aaryan | 1 year ago

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