Answer:
F)none of these
Step-by-step explanation:
Let the general equation of the circle be:
[tex]x^{2} +y^{2} +2ax+2by+c=0[/tex] where [tex]a,b,c[/tex] are constants
Conditions:
⇒[tex]x=0[/tex] and [tex]y=0[/tex] should satisfy the circle equation and should have only one solution to each case
put [tex]x=0[/tex] in circle equation, we get
[tex]y^{2} +2by+c=0[/tex], this equation has only one solution
⇒[tex]4b^{2} -4c=0[/tex]
⇒[tex]b=\sqrt{c} ,-\sqrt{c}[/tex]
similarly put [tex]y=0[/tex] in circle equation, we get
[tex]x^{2} +2ax+c=0[/tex], this equation has only one solution
⇒[tex]4a^{2} -4c=0[/tex]
⇒[tex]a=\sqrt{c} ,-\sqrt{c}[/tex]
⇒[tex]a,b[/tex] should be positive
⇒[tex]a=b=\sqrt{c}[/tex]
⇒ radius=2 and we know that[tex]r=\sqrt{a^{2}+b^{2} -c}[/tex]
⇒ [tex]2=\sqrt{c+c-c}[/tex]
⇒[tex]c=4[/tex]
∴[tex]a=b=2[/tex]
hence equation of the circle is:
[tex]x^{2} +y^{2} +4x+4y+4=0[/tex]
