Question

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

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

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A

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

B

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

C

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

D

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

we saw in the previous questions that it is better to start with the general equations and find the values of the variables g,f and c.we will use the given conditions to find them.Let the equations of the circle be
$x^2+y^2+2gx+2fy+c=0$
it is given that the circle touches the x-axis.
$\Rightarrow g^2=c$ - - - - - - - (1)
Length of the y-intercept is given as 6.
$\Rightarrow 2\sqrt{f^2-c}=6$
$\Rightarrow f^2-c=9$
$f^2=9+c$ - - - - - - - (2)
Next condition given is it touches the x-axis 3 uits from origin.so it can touch the x-axis at (3,0) or (-3,0),both being 3 units away from origin.
we will use relation (1) to replace 'c' with $g^2$.
$\Rightarrow x^2+y^2+2gx+2fy+g^2=0$ is the equation.
Let us find the point where it touches the x-axis.For that,put y=0.
$\Rightarrow x^2+2gx+g^2=0$
$\Rightarrow -g=\mp 3$
$\Rightarrow x=-g$
But we know that the x-coordinate of the point where the circle touches the x-axis is $\mp 3$
$\Rightarrow -g=\mp 3$
$\Rightarrow g=\mp 3$
$\Rightarrow c=g^2=9$
From (2),$f^2=9+c$
=9+9
=18
$\Rightarrow$ $f=\mp 3\sqrt{2}$
we got all the variables g,f and c.
$\Rightarrow$ we have $g=\mp 3,f=\mp 3\sqrt{2}andc=9$
$\Rightarrow$the equation of the circles are
$x^2+y^2\mp 6x6\sqrt{2}y+9=0$
$\Rightarrow$ All the options are correct.