Question

A circle touches both the axes and its centre lies in the
the axes and its centre lies in the fourth quadrant. If its radius is 1 then its
equation will be
(A) x2 + y2 - 2x +2y + 1 = 0
(B) x2 + y2 + 2x - 2y-1 = 0
(C) x2+ y2 - 2x - 2y + 1 = 0)
(D) x2 + y2 + 2x - 2y + 1 = 0​

Answer 1

Answer:

B because B is the answer...

It is because u cannot take

Answer 2

Since the circle touches both the coordinate axes hence there are two caese

Case 1: Let the center of circle (−a,−a) in third quadrant & the radius a Now, the equation of circle is (x+a)2+(y+a)2=a2

The center (−a,−a) will satisfy the equation of straight line: x−2y=3

⟹−a−2(−a)=3⟹a=3

Hence the equation of circle is: (x+3)2+(y+3)2=(3)2=9  

Case 2: Let the center of circle (a,−a) in fourth quadrant & the radius a Now, the equation of circle is (x−a)2+(y+a)2=a2

The center (a,−a) will satisfy the equation of straight line: x−2y=3

⟹a−2(−a)=3⟹a=1

Hence the equation of circle is:(x−1)2+(y+1)2=(1)2=1