Question

find the equation of circle which passed through origin and cuts off intercept of length 'a' ,each from positive direction of the axes

Answer 1

Here's the Answer..

Answer 2

equation of circle  (x - a/2)²   + (y - a/2)²  = (a/√2)²   or x²  + y² -ax - ay = 0

Step-by-step explanation:

circle passes through origin and cuts off intercept of length 'a' ,each from positive direction of the axes

These three points circle is passing through will be

(0 , a) , (0 , 0) , (a , 0)

x - axis & y - axis are perpendicular to each other hence angle will be 90 deg Between then hence line joining

(0 , a)  & (a , 0)  will be diameter

so center point of circle  =  (0 + a)/2 , (a + 0)/2  = a/2  , a/2

Diameter of circle = √a² + a² = a√2

Radius of circle = a√2/2  = a/√2

Hence equation of circle

(x - a/2)²   + (y - a/2)²  = (a/√2)²

=> x²  -ax + a²/4 + y² - ay + a²/4 = a²/2

=> x²  + y² -ax - ay = 0

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