Position of a Point With Respect to 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 center 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. The position of a point signifies that the point lies inside or outside the circle.

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In this article, we will cover the concept of the position of a point with respect to 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 three questions have been asked on this concept, including two in 2021, and one in 2022.

Position of a Point With Respect to Circle

A circle is the locus of a moving point such that its distance from a fixed point is constant.

The fixed point is called the centre (O) of the circle and the constant distance is called its radius (r)

Let $S$ be a circle and $P$ be any point in the plane. Then

$
S: x^2+y^2+2 g x+2 f y+c=0
$
Centre of the circle, $C(-g,-f)$

To check if the point $P\left(x_1, y_1\right)$ lies outside, on or inside the circle $S$
- If $C P^2-r^2>0$, then $C P$ is greater than $r$, which means $P$ lies outside the circle
- If $C P^2-r^2<0$, then $C P$ is lesser than $r$, which means $P$ lies inside the circle
- If $C P^2-r^2=0$, then $C P$ is equal to $r$, which means $P$ lies on the circle

Let us find the equation for $\mathrm{CP}^2-\mathrm{r}^2$

$
\begin{aligned}
& \left(x_1-(-g)\right)^2+\left(y_1-(-f)\right)^2-\left(g^2+f^2-c\right) \\
& x_1^2+g^2+2 g x_1+y_1^2+f^2+2 f y_1-\left(g^2+f^2-c\right) \\
& x_1^2+y_1^2+2 g x_1+2 f y_1+c
\end{aligned}
$
This expression can also be obtained by substituting the coordinates of point $P$ in the equation of the circle, and we call this expression $S_1$
So, if
(a) P lies outside the circle $\Leftrightarrow \mathrm{S}_1>0$
(b) P lies on the circle (on the circumference) $\Leftrightarrow \mathrm{S}_1=0$
(c) P lies inside the circle $\Leftrightarrow \mathrm{S}_1<0$

Greatest and Least Distance of a Point from a Circle
$\mathrm{S}_1$ be a circle and P be any point in the plane.

$
\begin{aligned}
& S_1: x^2+y^2+2 g x+2 f y+c=0 \\
& P=\left(x_1, y_1\right)
\end{aligned}
$
The centre of circle is $\mathrm{C}(-\mathrm{g},-\mathrm{f})$ and radius r is $\sqrt{g^2+f^2-c}$

(a) If P lies inside of the circle

The minimum distance of P from the circle = PA = AC - PC = r - PC

The maximum distance of P from the circle = PB = BC + PC = r + PC

(b) If P lies outside of the circle

The minimum distance of P from the circle = PA = CP - AC = CP - r

The maximum distance of P from the circle = PB = BC + PC = r + PC

Recommended Video Based on the Position of a Point with respect to a Circle

Solved Examples Based On the Position of a Point with respect to a Circle

Example 1: Which of the following points lie inside the circle $x^2+y^2+10 x-6 y-1=0$ ?
1) $(2,2)$
2) $(5,5)$
3) $(-5,-5)$
4) $(0,0)$

Solution:
As we have learned
Position of a point w.r.t a circle -
If $x_1^2+y_1^2+2 g x_1+2 f y_1+c<0$. then the point is inside the circle.
wherein the point is $P\left(x_1, y_1\right)$
If we put $(0,0)$
we get $0+0+0-0-1<0$.
Hence, the answer is the option 4.

Example 2: Which of the following circles has $(5,5)$ on it?
1) $x^2+y^2+25 x+25 y-625=0$
2) $x^2+y^2-5 x-5 y=0$
3) $x^2+y^2+5 x+5 y-50=0$
4) None of these

Solution:
As we have learned
Position of a point w.r.t a circle-
If $x_1^2+y_1^2+2 g x_1+2 f y_1+c=0$, then the point is on the circle.
Point is $P\left(x_1, y_1\right)=(5,5)$.
In the equation of the circle, $x=5, y=5$
If we substitute $(5,5)$, we get

$
5^2+5^2-5 \times 5-5 \times 5=50-50=0
$
Hence, the answer is the option 2.

Example 3: Which of the following points is outside the circle $x^2+y^2-6 x+10 y-10=0$ ?
1) $(2,2)$
2) $(5,5)$
3) $(3,3)$
4) All of the above

Solution:
As we learned
Position of a point w.r.t a circle
If $x_1^2+y_1^2+2 g x_1+2 f y_1+c>0$ then the point is outside the circle
Here,
For $(2,2)$
$S_1=4+4-12+20-10=6>0$, so it lies outside the circle
For $(5,5)$
$S_1=25+25-30+50-10=60>0$, so it lies outside the circle
For $(3,3)$
$S_1=9+9-18+30-10=20>0$, so it lies outside the circle
Hence, the answer is option 4.

Example 4: The set of values k , for which the circle $\mathrm{C}: 4 \mathrm{x}^2+4 \mathrm{y}^2-12 \mathrm{x}+8 \mathrm{y}+\mathrm{k}=0$ lies inside the fourth quadrant and the point $\left(1,-\frac{1}{3}\right)$ lies on or inside the circle C , is:
1) an empty set
2) $\left(6, \frac{65}{9}\right]$
3) $\left[\frac{80}{9}, 10\right)$
4) $\left(9, \frac{92}{9}\right]$

Solution:
Cirde $: x^2+y^2-3 x+2 y+\frac{k}{4}=0$
Centre : $\left(\frac{3}{2},-1\right)$ radius $: \sqrt{\left(\frac{3}{2}\right)^2+(1)^2-\frac{\mathrm{k}}{4}}$
The radius must be less than 1 for the circle to lie in the fourth quadrant.

$
\Rightarrow \quad \frac{9}{4}+1-\frac{k}{4}<1 \Rightarrow \mathrm{k}>9
$

The point $\left(1,-\frac{1}{3}\right)$ lies on or inside the circle

$
\begin{aligned}
& \Rightarrow \mathrm{s}_1 \leqslant 0 \Rightarrow 1+\frac{1}{9}-3-\frac{2}{3}+\frac{\mathrm{k}}{4} \leqslant 0 \Rightarrow \frac{\mathrm{k}}{4} \leqslant \frac{8}{3}-\frac{1}{9}=\frac{23}{9} \\
& \Rightarrow \mathrm{k} \leqslant \frac{92}{9}
\end{aligned}
$
From (1) and (2)

$
\mathrm{k} \in\left(9, \frac{92}{9}\right]
$
Hence, the answer is the option (4).

Example 5: Let the circle $S: 36 x^2+36 y^2-108 x+120 y+C=0$ be such that it neither intersects nor touches the co-ordinate axes. If the point of intersection of the lines, $x-2 y=4$ and $2 x-y=5$ lies inside the $\underline{c}$ circle $S$, then
1) $\frac{25}{9}<C<\frac{13}{3}$
2) $100<C<165$
3) $81<\mathrm{C}<156$
4) $100<\mathrm{C}<156$

Solution:

$
\begin{aligned}
& 36 x^2+36 y^2-108 x+120 y+C=0 \\
& \Rightarrow x^2+y^2-3 x+\frac{10}{3} y+\frac{C}{36}=0
\end{aligned}
$
As there is no real $x$ or $y_{\text {intercept, }}$,

$
\begin{aligned}
& \therefore 2 \sqrt{g^2-c} \& 2 \sqrt{f^2-c} \text { are not real } \\
& \Rightarrow g^2-c<0 \text { and } f^2-c<0 \\
& \Rightarrow \frac{9}{4}-\frac{C}{36}<0 \text { and } \frac{25}{9}-\frac{C}{36}<0 \\
& \Rightarrow C>81 \text { and } C>100 \Rightarrow C>100
\end{aligned}
$

Point of intersection of $2 x-y=5$ and $x-2 y=4$ is $(2,-1)$
This lies inside the circle S.

$
\begin{aligned}
& \therefore S(2,-1)<0 \\
& \Rightarrow 2^2+(-1)^2-3(2)+\frac{10}{3}(-1)+\frac{C}{36}<0 \\
& \Rightarrow C<156
\end{aligned}
$
So $100<C<156$
Hence, the answer is the option (4).