Intercepts on the Axes made by a Circle

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 Forms of a Circle

When the circle touches X-axis

$(a, b)$ be the centre of the circle, then radius $=|b|$
$\therefore$ equation of circle becomes

$
\begin{aligned}
& \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

$(a, b)$ be the centre of the circle, then radius $=|a|$
$\therefore$ equation of circle becomes

$
\begin{aligned}
& \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=0
\end{aligned}
$

When the circle touches both the axes:

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

Note:

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

Solved Examples Based on Intercept made by Circles on Axis

Example 1: The x and y intercepts of the circle $(x-1)^2+(y-1)^2=4$ respectively are
Solution:

$
\begin{aligned}
& (x-1)^2+(y-1)^2=4 \\
& x^2+y^2-2 x-2 y-2=0
\end{aligned}
$
So, $g=-1, f=-1, c=-2$.

$
\begin{aligned}
& x \text {-intercept }=2 \sqrt{g^2-c}=2 \sqrt{(-1)^2-(-2)}=2 \sqrt{3} \\
& \text { y-intercept }=2 \sqrt{f^2-c}=2 \sqrt{(-1)^2-(-2)}=2 \sqrt{3}
\end{aligned}
$
Example 2: Find the number of points where the circle $x^2+y^2+4 x-4 y+7=0$ intersects the x -axis.
Solution:
Intercepts Made by Circle on the Axis:

$
\text { For } x^2+y^2+2 g x+2 f y+c=0
$
Length of $x$-intercept: $2 \sqrt{\mathrm{g}^2-\mathrm{c}}$
Now

$
\begin{aligned}
& x^2+y^2+4 x-4 y+7=0 \\
& x-\text { intercept }=2 \sqrt{g^2-c}=2 \sqrt{(2)^2-7}=2 \sqrt{-3}
\end{aligned}
$
As this is not a real number, so there is no $x$-intercept made by this circle with the x -axis, and hence there is no point of intersection of this circle with the x -axis

Example 3: If the circle $x^2+y^2-2 g x+6 y-19 c=0, g, c \in \mathbb{R}_{\text {passes through the point }}(6,1)$ and its centre lies on the linex $-2 c y=8$, then the length of intercept made by the circle on $x-a x i s$ is
Solution:
Given circle $x^2+y^2-29 x+6 y-19 c=0$
passes through $(6,1)$

$
12 \mathrm{~g}+19 \mathrm{c}=43 \cdots
$
Centre $(\mathrm{g},-3$ ) lies on the given line
So, $g+6 c=8 \cdots(2)$
Solve equation (1) \& (2)

$
c=1 \& y=2
$

equation of circle $x^2+y^2-4 x+6 y-19=0$
Length of intercept on the $x$-axis

$
=2 \sqrt{\mathrm{g}^2-\mathrm{c}}=2 \sqrt{23}
$

Example 4: If $\mathrm{P}(2,8)$ is an interior point of a circle $\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}+4 \mathrm{y}-\mathrm{p}=0$ which neither touches nor intersects the axes, then set for p is:
Solution:
For internal point $\mathrm{p}(2,8) 4+64-4+32-\mathrm{p}<0 \Rightarrow \mathrm{p}>96$ and x -intercept $=2 \sqrt{1+\mathrm{p}}$ therefore $1+\mathrm{p}<0$
$\Rightarrow \mathrm{p}<-1$ and y -intercept $=2 \sqrt{4+\mathrm{p}} \Rightarrow \mathrm{p}<-4$

Example 5: A variable circle passes through the fixed point $A(p, q)$ and touches the x -axis. The locus of the other end of the diameter through $A$ is
Solution:
Let the other diametric end be $\mathrm{P}(\mathrm{h}, \mathrm{k})$
So centre is $\left(\frac{p+h}{2}, \frac{q+R}{2}\right)$
Radius $=\sqrt{\left(\frac{h-p}{2}\right)^2+\left(\frac{k-q}{2}\right)^2}$
For a circle touching the $x$-axis, radius $=\left(\frac{q+k}{2}\right)$
So $\left(\frac{h-p}{2}\right)^2+\left(\frac{k-q}{2}\right)^2=\left(\frac{k+q}{2}\right)^2$
we get $(h-p)^2=4 k g$
i.e. $(x-p)^2=4 q y$, a parabola

Summary

The circles are foundational shapes with unique properties and applications in various mathematics, science, and engineering fields. Understanding the properties, equations, and applications of circles is essential for solving geometric problems, designing objects, and analyzing natural phenomena.