Question

Find the equation of a circle which passes through origin and cuts off intercepts a and b
on the coordinate axes.​

Answer 1

Answer:

Answer: general equation of a circle:x^2+y^2+2gx+2fy+c=0: points lie on the circle are (0,. 0)(a,0) and (0,b). and,B be a point (b,0) at X axis.

Answer 2

Step-by-step explanation:

General equation of circle x2+y2+2gx+2fy+c=0

Points lie on the circle are (0,0);(a,0) & (0,b).

by putting (0,0) we get c = 0 ; 

by putting (a,0) we get a2+2ga=0

by putting (0,b) we get b2+2fb=0

Hence a=−2g;b=−2f

∴ equation of circle will be x2+y2−ax−by=0