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Circle - Practice Questions & MCQ

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

Quick Facts

  • Circle(Definition) is considered one the most difficult concept.

  • 196 Questions around this concept.

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The equation of a circle with origin as a centre and passing through equilateral triangle whose median is of length $3 a$ is

A circle touches the $x$-axis and also touches the circle with centre at $(0,3)$ and radius 2. The locus of the centre of the circle is

The point diametrically opposite to the point $P(1,0)$ on the circle $x^2+y^2+2 x+4 y-3=0$ is

 

 

If a circle of radius R passes through the origin O and intersects the coordinate axes at A and B, then the locus of the foot of perpendicular from O on AB is : 

Let $S_1: x^2+y^2=9$ and $S_2:(x-2)^2+y^2=1$. Then the locus of the centre of a variable circle S which touches $S_1$ internally and $S_2$ externally always passes through the points:

Let C be the centre of the circle $\mathrm{x}^2+\mathrm{y}^2-\mathrm{x}+2 \mathrm{y}=\frac{11}{4}$ and P be a point on the circle. A line passes through the point C , makes an angle of $\frac{\pi}{4}$ with the line CP and intersects the circle at the points $\mathrm{Q} {\text {and }} \mathrm{R}$. Then the area of the triangle $\mathrm{PQR}{\text {(in unit }}{ }^2$ ) is :

A square is inscribed in the circle \mathrm{x^{2}+y^{2}-2 x+4 y+3=0}. Its sides are parallel to the coordinate axes. Then one vertex of the square is:

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The equation \mathrm{x^{2}+y^{2}+4 x+6 y+13=0} represents:

Equation of curve for which normal at its any point pass through (0,0)

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$
\text { Find the equation of a circle touching } X \text {-axis and having center }(1,-2)
$

 

Concepts Covered - 1

Circle(Definition)

Circle

Definition

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 constant distance is called its radius (r)

Equation of circle

Centre-Radius Form

The equation of a circle with centre at $C(h, k)$ and radius $r$ is $(x-h)^2+(y-k)^2=r^2$

Let $P(x, y)$ be any point on the circle. Then, by definition, $|C P|=r$.
Using the distance formula, we have

$
\sqrt{(\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2}=\mathrm{r}
$

i.e.

$
(\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2=\mathrm{r}^2
$
If the centre of the circle is the origin or $(0,0)$ then equation of the circle becomes

$
\begin{aligned}
& (x-0)^2+(y-0)^2=r^2 \\
& \text { i.e. } x^2+y^2=r^2
\end{aligned}
$
General Form:
The equation of a circle with centre at $(\mathrm{h}, \mathrm{k})$ and radius r is

$
\begin{aligned}
& \Rightarrow(x-h)^2+(y-k)^2=r^2 \\
& \Rightarrow x^2+y^2-2 h x-2 k y+h^2+k^2-r^2=0
\end{aligned}
$
Which is of the form :

$
\mathbf{x}^2+\mathrm{y}^2+2 \mathrm{gx}+2 \mathbf{y}+\mathbf{c}=\mathbf{0}
$
This is known as the general equation of the circle.
To get radius and centre if only the equation of the circle (ii) is given:
Compare eq (i) and eq (ii)

$
\mathrm{h}=-\mathrm{g}, \mathrm{k}=-\mathrm{h} \text { and } \mathrm{c}=\mathrm{h}^2+\mathrm{k}^2-\mathrm{r}^2
$
Coordinates of the centre $(-g,-\mathrm{f})$
Radius $=\sqrt{g^2+f^2-c}$

Nature of the Circle
For the standard equation of a circle $\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0$ whose radius is given as $\sqrt{g^2+f^2-c}$
Now the following cases arise
1. If $g^2+f^2-c>0$, then the radius of the circle will be real. Hence, the circle is a real circle.
2. If $g^2+f^2-c=0$, then the radius of the circle will be real $(=0)$. Hence, the circle is a Point circle because the radius is 0 .
3. If $g^2+f^2-c<0$, then the radius of the circle will be imaginary. Hence, the circle is an imaginary circle.

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Circle(Definition)

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Circle(Definition)

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

Page No. : 4.1

Line : 1

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