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# Line and Circle - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• Line and Circle is considered one the most difficult concept.

• 136 Questions around this concept.

## Solve by difficulty

The equation of the circle of radius $\sqrt{2}$ which touches the line $x+y=1 at \: (2,-1)$ is:

Let $L 1$ be a straight line passing through the origin and $L 2$ be the straight line  $\mathrm{X+Y=1}$. If the intercepts made by the circle $\mathrm{x 2+y 2-x+3 y=0}$ on L1 and L2 are equal, then which of the following equations can represent L1 :

If $\mathrm{\frac{x}{a}+\frac{y}{b}=1}$ touches the circle  $\mathrm{x^{2}+y^{2}=r^{2}}$, then prove that the point $\mathrm{\left(\frac{1}{a}, \frac{1}{b}\right)}$ lies.

Image of the circle $\mathrm{x^{2}+y^{2}-6 x+8=0}$ in the line $\mathrm{y=x}$ is

## Concepts Covered - 1

Line and Circle

Line and Circle

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

$\begin{array}{l}{\text { Equation of circle } \mathrm{S} : \mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{a}^{2} \quad \ldots(\mathrm{i})} \\ {\text { Equation of line } \mathrm{L} : \mathrm{y}=\mathrm{mx}+\mathrm{c}} \\ {\text {To find their point(s) of intersection, we can solve these equations simultaneously }} \\ {\mathrm{x}^{2}+(\mathrm{mx}+\mathrm{c})^{2}=\mathrm{a}^{2}} \\ {\left(1+\mathrm{m}^{2}\right) \mathrm{x}^{2}+2 \mathrm{mcx}+\mathrm{c}^{2}-\mathrm{a}^{2}=0 \quad \ldots \text { (iii) }}\end{array}$

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

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

In this case the line is a 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, Discriminant of equation (iii) < 0

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

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

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

## Study it with Videos

Line and Circle

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## Books

### Reference Books

#### Line and Circle

Mathematics for Joint Entrance Examination JEE (Advanced) : Coordinate Geometry

Page No. : 4.14

Line : 22