Intercepts Made by Circle on the Axis - Practice Questions & MCQ

Intercepts Made by Circle on the Axis 

If the equation of Circle is $x^2+y^2+2 g x+2 f y+c=0$, then
Length of $x$-intercept $: 2 \sqrt{g^2-c}$
Length of y -intercept : $2 \sqrt{\mathrm{f}^2-\mathrm{c}}$

Proof:

from the figure
length of intercepts on $X-$ axis and $Y-$ axis are $|A B|$ and $|C D|$

$
|A B|=\left|x_2-x_1\right|,|C D|=\left|y_2-y_1\right|
$
Put $y=0$, to get points A and B , where circle intersects the $X-$ axis

$
\Rightarrow x^2+2 g x+c=0
$
Since, circle intersects $X-$ axis at two points $A\left(x_1, 0\right)$ and $B\left(x_2, 0\right)$ so x 1 and x 2 are roots of the above equation, and hence, $x_1+x_2=-2 g x, x_1 x_2=c$

$
|A B|=\left|x_2-x_1\right|=\sqrt{\left(x_2+x_1\right)^2-4 x_1 x_2}=2 \sqrt{g^2-c}
$
Similarly,

$
|C D|=2 \sqrt{f^2-c}
$
 

Different Form of a Circle

When the circle touches X-axis

$\begin{aligned} & (\mathrm{a}, \mathrm{b}) \text { be the centre of the circle, then radius }=|\mathrm{b}| \\ & \therefore \text { equation of circle becomes } \\ & \Rightarrow(\mathrm{x}-\mathrm{a})^2+(\mathrm{y}-\mathrm{b})^2=\mathrm{b}^2 \\ & \Rightarrow \mathrm{x}^2+\mathrm{y}^2-2 \mathrm{ax}-2 \mathrm{by}+\mathrm{a}^2=0\end{aligned}$

When the circle touches Y-axis

$\begin{aligned} & (\mathrm{a}, \mathrm{b}) \text { be the centre of the circle, then radius }=|\mathrm{a}| \\ & \therefore \text { equation of circle becomes } \\ & \Rightarrow(\mathrm{x}-\mathrm{a})^2+(\mathrm{y}-\mathrm{b})^2=\mathrm{a}^2 \\ & \Rightarrow \mathrm{x}^2+\mathrm{y}^2-2 \mathrm{ax}-2 \mathrm{by}+\mathrm{b}^2=\mathbf{0}\end{aligned}$

When the circle touches both the axes:

$(a, a)$ be the centre of the circle, then radius $=|a| \mid$

$
\begin{aligned}
& \therefore \text { equation of circle becomes } \\
& \Rightarrow(\mathrm{x}-\mathrm{a})^2+(\mathrm{y}-\mathrm{a})^2=\mathrm{a}^2 \\
& \Rightarrow \mathrm{x}^2+\mathrm{y}^2-2 \mathrm{ax}-2 \mathrm{ay}+\mathrm{a}^2=0
\end{aligned}
$

Note:

In this case, the centre can also be $(a,-a)$ and radius $|a|$.