Question

Find the equation to the circle which touches the axis of $x$ at a distance $3$ from the origin and intercepts a distance $6$ on the axis of $y$.

• A. $x^2+y^2-3x-3\sqrt{2}y+5=0$
• B. $x^2+y^2-6x-3\sqrt{2}y+9=0$
• C. $x^2+y^2-3x+3\sqrt{2}y+4=0$
• D. $x^2+y^2-4x-5\sqrt{2}y-10=0$

Answer 1

The correct option is B $x^2+y^2-6x-3\sqrt{2}y+9=0$

The equation of the circle with centre $(h,k)$ and radius $a$ is

$(x-h{)}^2+(y-k{)}^2=a^2$

When the circle touches the x-axis the ordinate of the centre is equal to the

radius of the circle i.e. $k=a$.

Therefore, the equation becomes

$(x-h{)}^2+(y-a{)}^2=a^2$

$x^2+y^2-2hx-2ay+h^2=0$

The circle passes through$(3,0)$ and the intercept made by a circle with

$y$-axis is $2\sqrt{f^2-c}$

So, $2\sqrt{4a^2-h^2}=6$ ......(1)

and $9-6h+h^2=0$ ......(2)

Solving (1) and (2), we get

$h=3$

$a=\frac{3}{\sqrt{2}}$

So, the equation becomes

$x^2+y^2-6x-3\sqrt{2}y+9=0$