Question

An equation of a circle which touches the y-axis at $(0,2)$ and cuts off an intercept $3$ from the x-axis is

• A. $x^2+y^2+4x-5y+4=0$
• B. $x^2+y^2+5x-4y+4=0$
• C. $x^2+y^2-5x-4y+4=0$
• D. $x^2+y^2-5x+4y+4=0$

Answer 1

Answer: (B), (C)

$x^2+y^2+5x-4y+4=0$
$x^2+y^2-5x-4y+4=0$

As the required circle touches the y-axis at $(0,2)$
let its equation be $(x-\alpha {)}^2+(y-2{)}^2={\alpha }^2$ or $x^2+y^2-2\alpha x-4y+4=0$
This circle meets the x-axis at the points where $x^2-2\alpha x+4=0$,
which gives two values of $x$, say $x_1+x_2,$
such that $x_1+x^2=2\alpha$ and $x_1{x}_2=4.$
Now, we are given that $|x_1-x_2|=3.$
$\therefore (x_1+x_2{)}^2-4x_1{x}_2=(x_1-x_2{)}^2=9$
$\Rightarrow 4{\alpha }^2-16=9\Rightarrow {\alpha }^2=\frac{25}{4}\Rightarrow \alpha =\pm \frac{5}{2}$
Hence the equation of the required circle is $x^2+y^2\pm 5x-4y+4=0$