Question

The equation of the circle touching the $y-$axis at $(0,3)$ and making an intercept of $8$ on the $x-$axis is

• A. $x^2+y^2+10x-6y+9=0$
• B. $x^2-y^2-10x-6y-9=0$
• C. $x^2+y^2-10x-6y+9=0$
• D. $x^2+y^2+10x+6y+9=0$

Answer 1

Answer: (A), (C)

$x^2+y^2-10x-6y+9=0$
$x^2+y^2+10x-6y+9=0$

Using the above information,
Circle touching a line $L=0$ at a point $(x_1.y_1)$ on it is
${(x-x_1)}^2+{(y-y_1)}^2+\lambda L=0,\lambda \in R$
We have the circle as ${(x-0)}^2+{(y-3)}^2+\lambda x=0,y=0$
$\Rightarrow x^2+y^2-6y+9-\lambda x=0$
As $y=0,$ the above equation gets reduced to
$x^2+\lambda x+9=0$
$\Rightarrow x=x_1,x=x_2$say, where $x_1+x_2=-\lambda ,x_1{x}_2=9$
$|x_1-x_2|=8\Rightarrow {(x_1-x_2)}^2=64$
${(x_1+x_2)}^2-4x_1{x}_2=64$
$\Rightarrow {(-\lambda )}^2-4\times 9=64$
$\Rightarrow \lambda =\pm 10$
The circles are $x^2+y^2\pm 10x-6y+9=0$