Question

If a circle touches x axis at $(3,0)$ and passing through the point $(2,1)$ then its equation is

• A. $x^2+y^2-6x+2y+9=0$
• B. $x^2+y^2-6x-2y+9=0$
• C. $x^2+y^2-6x-2y-9=0$
• D. $x^2+y^2-6x+2y-9=0$

Answer 1

Answer: (B)

$x^2+y^2-6x-2y+9=0$

Centre of circle which touches x-axis at $(3,0)$ is at $(3,k)$ and radius of circle is $k,$
$\therefore (x-3{)}^2+(y-k{)}^2=k^2$
$x^2+y^2-6x+9-2ky=0$
This circle passes through (2,1),
$5-12+9-2\times 5\times k\times 1=0$
$2k=2$
$k=1$
Hence equation of circle is
$\Rightarrow x^2+y^2-6x-2y+9=0$