Intercepts on the Axes made by a Circle

A circle is one of the most fundamental geometric shapes, consisting of all points in a plane that is equidistant from a fixed point called the centre of a circle. It is a very basic shape that is constantly used in mathematics. The main applications of the circle are in geometry, engineering for designing circular instruments, physics, and technology.

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This Story also Contains

1. Intercepts Made by Circle on the Axis
2. Different Forms of a Circle
3. Solved Examples Based on Intercept made by Circles on Axis
4. Summary

In this article, we will cover the concept of the intercept made by a circle. This concept falls under the broader category of coordinate geometry. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of sixteen questions have been asked on this concept, including one in 2013, one in 2014, one in 2016, eight in 2021, two in 2022, and two in 2023.

Intercepts Made by Circle on the Axis

If the equation of Circle is $x^2+y^2+2gx+2fy+c=0$, 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 $|AB|$ and $|CD|$

$|AB|=|x_2-x_1|,|CD|=|y_2-y_1|$
Put $y=0$, to get points A and B , where circle intersects the $X-$ axis

$\Rightarrow x^2+2gx+c=0$
Since, circle intersects $X$ - axis at two points $A(x_1,0)$ and $B(x_2,0)$ so x 1 and x 2 are roots of the above equation, and hence, $x_1+x_2=-2gx,x_1{x}_2=c$

$|AB|=|x_2-x_1|=\sqrt{{(x_2+x_1)}^2-4x_1{x}_2}=2\sqrt{g^2-c}$
Similarly,

$|CD|=2\sqrt{f^2-c}$

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

$\begin{array}{rl}& \Rightarrow (\mathrm{x}-\mathrm{a}{)}^2+(\mathrm{y}-\mathrm{b}{)}^2={\mathrm{b}}^2\\ & \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{ax}-2\mathrm{by}+{\mathrm{a}}^2=0\end{array}$

When the circle touches Y-axis

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

$\begin{array}{rl}& \Rightarrow (\mathrm{x}-\mathrm{a}{)}^2+(\mathrm{y}-\mathrm{b}{)}^2={\mathrm{a}}^2\\ & \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{ax}-2\mathrm{by}+{\mathrm{b}}^2=0\end{array}$

When the circle touches both the axes:

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

Note:

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

Recommended Video Based on Intercept made by Circles on Axis

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:

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

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

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

$\begin{array}{rl}& x^2+y^2+4x-4y+7=0\\ & x-\text{intercept}=2\sqrt{g^2-c}=2\sqrt{(2{)}^2-7}=2\sqrt{-3}\end{array}$
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 $x^2+y^2-2gx+6y-19c=0,g,c\in {\mathbb{R}}_{\text{passes through the point}}(6,1)$ and its centre lies on the linex $-2cy=8$, then the length of intercept made by the circle on $x-axis$ is
Solution:
Given circle $x^2+y^2-29x+6y-19c=0$
passes through $(6,1)$

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

$c=1\mathrm{\&}y=2$

equation of circle $x^2+y^2-4x+6y-19=0$
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 $1+\mathrm{p}<0$界
$\Rightarrow \mathrm{p}<-1$ 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 $(\frac{p+h}{2},\frac{q+R}{2})$
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 $(h-p{)}^2=4kg$
i.e. $(x-p{)}^2=4qy$, 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.