Compute the centre and radius of the circle 2x2 + 2y2 – x = 0
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Solution:
Given that, the circle equation is 2x2 + 2y2 – x = 0
This can be written as:
⇒ (2x2-x) + y2 = 0
⇒ 2{[x2 – (x/2)] +y2} = 0
⇒{ x2 – 2x(¼) + (¼)2} +y2 – (¼)2 = 0
Now, simplify the above form, we get
⇒(x- (¼))2 + (y-0)2 = (¼)2
The above equation is of the form (x – h)2 + (y – k)2= r2
Therefore, by comparing the general form and the equation obtained, we can say
h= ¼ , k = 0, and r = ¼.
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