Intersection of Line and 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 interaction between a line and a circle in a plane can result in different scenarios based on their relative positions. The main applications of the circle are in geometry, engineering for designing circular instruments, physics, and technology.

In this article, we will cover the concept of the line and 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 nine questions have been asked on this concept, including one in 2013, one in 2019, two in 2020, one in 2021, one in 2022, and three in 2023.

Line and Circle

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

S is a circle with centre O and radius r, and L is a straight line in the plane of the circle.

Equation of circle $S: x^2+y^2=a^2 \quad \ldots$ (i)
Equation of line $L: y=m x+c$
To find their point(s) of intersection, we can solve these equations simultaneously

$\begin{aligned}
& x^2+(m x+c)^2=a^2 \\
& \left(1+m^2\right) x^2+2 m c x+c^2-a^2=0
\end{aligned}$

Case (1)

If line L intersects the circle S in two distinct points, then Equation (iii) will have two real and distinct roots

So, Discriminant of equation (iii) > 0

In this case, the line is a secant to the circle, i.e., it represents the equation of chord to the circle.

Case (2)

If the line L touches the circle, then Equation (iii) will have two equal real roots

So, Discriminant of equation (iii) = 0

$\begin{aligned} & \mathrm{B}^2-4 \mathrm{AC}=0 \\ & 4 \mathrm{~m}^2 \mathrm{c}^2-4\left(1+\mathrm{m}^2\right)\left(\mathrm{c}^2-\mathrm{a}^2\right)=0 \\ & \mathrm{a}^2=\frac{\mathrm{c}^2}{1+\mathrm{m}^2} \\ & \mathrm{c}^2=a^2\left(1+\mathrm{m}^2\right)\end{aligned}$

In this case, the line is tangent to the circle

This is also the condition of tangency to the circle.

Case (3)

If the line L and the circle S have no common points, then Equation (iii) will have imaginary roots

So, the Discriminant of equation (iii) < 0

Alternate method to check the position of the line with respect to the circle

We can find the distance OM between the line and the center of the circle, and compare it with the radius of the circle

• If OM < r, then the line is a chord (secant) to the circle
• If OM = r, then the line is tangent to the circle
• If OM > r, then the line does not intersect the circle at any point

Recommended Video Based on Intersection of Line and Circle

Solved Examples Based on Intersection of Line and Circle

Example 1: The circle $x^2+y^2=4 x+8 y+5$ intersects the line $3 x-4 y=m$ at two distinct points if
1) $-85<m<-35$
2) $-35<m<15$
3) $15<m<65$
4) $35<m<85$

Solution:
As we learned in
Condition of tangency -

$c^2=a^2\left(1+m^2\right)$

- wherein

If $y=m x+c$ is a tangent to the circle $x^2+y^2=a^2$ $\Phi_2^2+y^2-4 x-8 y-5=0$
and $3 x-4 y=m$

$\begin{aligned}
& x=\frac{m+4 y}{3} \\
& \frac{(m+4 y)^2}{9}+y^2-\frac{(4 m+16 y)}{3}-8 y-5 \Rightarrow 0 \\
& \frac{25}{9} y^2+y\left(\frac{8 m}{9}-\frac{40}{3}\right)+\left(\frac{m^2}{9}-5\right) \Rightarrow 0 \\
& 25 y^2+8 y(m-120)+\left(m^2-45\right) \Rightarrow 0 \\
& D>0 \\
& 8^2(m-120)^2-4 \times 25\left(m^2-45\right) \Rightarrow 0
\end{aligned}$
$\begin{aligned}
& 16 m^2-32 \times 120 m+16 \times 120^2-25 m^2+25 \times 45>0 \\
& -9 m^2-32 \times 120 m+16 \times 120^2+25 \times 45>0
\end{aligned}$

On solving, we get

$\begin{aligned}
& (m+35)(m-15)<0 \\
& m \epsilon(-35,15)
\end{aligned}$

Hence, the answer is the option 2.

Example 2: Two common tangents to the circle $x^2+y^2=2 a^2$ and parabola $y^2=8 a x$ are
1) $x= \pm(y+2 a)$
2) $y= \pm(x+2 a)$
3) $x= \pm(y+a)$
4) $y= \pm(x+a)$

Solution:
Equation of tangent - of parabola $y^2=4 a x$
$y=m x+\frac{a}{m}$
and
If $y=m x+c$ is tangent to the circle $x^2+y^2=a^2$
Condition of tangency - $c^2=a^2\left(1+m^2\right)$
Tangent to circle $x^2+y^2=2 a^2$
is $y=m x+a \sqrt{2\left(1+m^2\right)}$

and Tangent to parabola y²=8ax
$\text { is } y=m x+\frac{2 a}{m}$we get $a \sqrt{2\left(1+m^2\right)}=\frac{2 a}{m}$

$\begin{aligned}
& \left(m^2+1\right) \cdot m^2=2 \\
& m^4+2 m^2-m^2-2=0 \\
& m^2=1 ; m= \pm 1
\end{aligned}$

we get $y= \pm(x+2 a)$
Hence the answer is the option (2).

Example 3: If the chord $y=m x+1$ of the circle $x^2+y^2=1$ subtends an angle of measure $45^{\circ}$ at the major segment of the circle then the value of $m$ is
Solution:
As we learned in
Condition of tangency -

$c^2=a^2\left(1+m^2\right)$
It $y=m x+c$ is a tangent to the circle $x^2+y^2=a^2$
Equation of circle $\mathrm{x}^2+\mathrm{y}^2=1$
Now, $y=m x+1$

subtends 45⁰ at the major segment of the circle then, it will subtend angle 90⁰ at the origin.

making $x^2+y^2=1$ homogeneous in the second degree with the help of $y=m x+1$

So, $x^2+y^2=(y-m x)^2$

$x^2\left(1-m^2\right)+2 m x y=0$

for $a x^2+b y^2+2 h x y+2 g x+2 f y+c=0$ to subtend $90^0$ at origin it should satisfy $a+b=0$
On solving $m= \pm 1$

Example 4: If the line $3 x-4 y-\lambda=0$ touches the circle $x^2+y^2-4 x-8 y-5=0$ at ( $\mathrm{a}, \mathrm{b}$ ), then $\lambda+a+b$ is equal to:
Solution:
Since the given line touches the given circle, the length of the perpendicular from the centre $(2,4)$ of the circle from the line $3 x-4 y-\lambda=0$ is equal to the radius $\sqrt{(4+16+5)}=5$ of the circle.

$\Rightarrow \frac{(3 \times 2-4 \times 4-\lambda)}{\sqrt{(9+16)}}= \pm 5 \Rightarrow \lambda=15 \text { or }-35$
Now, the equation of the tangent at ( $a, b$ ) to the given circle is

$\begin{gathered}
x a+y b-2(x+a)-4(y+b)-5=0 \\
\Rightarrow(a-2) x+(b-4) y-(2 a+4 b+5)=0
\end{gathered}$
If it represents the given line $3 x-4 y-\lambda=0$, then

$\begin{aligned}
& \frac{a-2}{3}=\frac{b-4}{-4}=\frac{2 a+4 b+5}{\lambda}=l \text { (say) } \\
& \text { Then } a=3 l+2, b=4-4 l \text { and } 2 a+4 b+5=\lambda l \\
& \Rightarrow \quad 2(3 l+2)+4(4-4 l)+5=15 l(\text { if } \lambda=15)
\end{aligned}$

$\begin{aligned}
& \Rightarrow \quad 2(3 l+2)+4(4-4 l)+5=15 l(\text { if } \lambda=15) \\
& \Rightarrow \quad l=1 \Rightarrow a=5, b=0 \text { and } \lambda+a+b=20
\end{aligned}$
Again, if $\lambda=-35$
(from i)

$25-10 l=-35 l \Rightarrow l=-1 \Rightarrow a=-1, b=8$

and $\lambda+a+b=-35-1+8=-28$
Example 5: The points of intersection of the line $a x+b y=0,(a \neq b)$ and the circle $x^2+y^2-2 x=0$ are $\mathrm{A}(\alpha, 0)$ and $\mathrm{B}(1, \beta)$. The image of the circle with AB as a diameter in the line $x+y+2=0$ is :
1) $x^2+y^2+3 x+3 y+4=0$

$\text { 2x } x^2+y^2+3 x+5 y+8=0$

3) $x^2+y^2-5 x-5 y+12=0$
4) $x^2+y^2+5 x+5 y+12=0$

Solution:
Only possibilities is $\alpha=0, \beta=1$
Equation of circle

$\begin{aligned}
& (x-0)(x-1)+(y-0)(y-1)=0 \\
& x^2+y^2-x-y=0
\end{aligned}$
Image of the circle in line $x+y+2=0$ is $x^2+y^2+5 x+5 y+12=0$
Hence, the answer is the option 4.