Standard equation of a circle: Formulas, Examples and Important Questions

A circle is made when a set of points on the same plane are placed at an equal distance from the centre, then if we join all the points, it will form a circle.

Table of Content

Formulas
Examples
Important Topics for JEE Main
Sample Questions

Also read: Quadrilateral Formula

Formulas

[Click Here for Sample Questions]

First, let’s discuss What exactly is a circle or how a circle forms. A circle is nothing but a collection of points. But can random points form a circle? Absolutely “NO”.

{r – Radius of a circle, d – Diameter of a circle}

“r” is the radius of a determined length in the circle and if we know one radius, any radius which comes from the centre of a circle to the circumference of a circle are of the same length i.e., “r”. And if we extend the radius through the centre of the circle to the other edge of the circle, that distance is called the diameter of circle “d”, whose length is twice the length of the radius.

And these basic properties apply to all circles and all sizes.

Now, as we know Mathematics has several branches like Geometry, Algebra, logic, Number theory etc.

Till now, we have seen the geometrical representation of Circle, i.e.

{r – Radius of a circle, d – Diameter of a circle}

Discover about the Chapter video:

Conic Sections Detailed Video Explanation:

Now, let’s try to understand the algebraic representation of a circle.

And an algebraic representation of any geometric figure is called an Equation.

To find the Algebraic expression of a circle, we need to find its equation. And to find the equation of a circle, let’s name a few things. Let the centre be O and its coordinates be (h, k). Similarly, let P be any point on a circle with (x, y) as its coordinates. Also, let’s denote the radius by r, So the distance between the point O and P is nothing but the radius.

ikthe distance between the point O and P is nothing but the radius — Distance between the point O and P is nothing but the radius

Now, how do we find the distance between the points O and P, i.e. OP.

As we know the coordinates of point O and P, we can find the distance OP using the distance formula

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

And after equating both the side we get this

\((x - h)^2 + (y - k)^2 = r^2 \)

This is the general equation or the standard form of the circle.

Now let's try to find the equation of a circle whose centre lies on the origin O, Since the centre is the origin, the value of (h, k) = (0,0).

Now, if we substitute these values in these equations:

\((x - h)^2 + (y - k)^2 = r^2 \)

We get

(x – 0)² + (y – 0)² = r²

(x)² + (y)² = r²

And thus, we get the equation of a circle with origin at its centre.

Now we are going to be looking at Circle at the coordinate plane, and applying the standard form of the equation of circle.

The information which we need is the location of the centre of the circle and the length of the radius. Let’s take (h, k) in the formula to represent the x and y coordinates of the centre point. And the letter r just represents the length of the radius of the circle.

Now we are going to solve some examples using the standard form of the equation of a circle.

Example 1) So for our first example, we want to write our standard form of equation

\((x - h)^2 + (y - k)^2 = r^2 \)

for circle “S” as shown in the graph with centre at (4,3)

First, let’s write the standard form of equation of a circle

\((x - h)^2 + (y - k)^2 = r^2 \)

Now, this just means that I need to find the centre of the circle and the length of the radius.

Now question already told me that at point S is the centre with coordinate (4,3)

Now to find the radius, I start with the centre point and go till the circumference of the circle, and count the unit. i.e., 3. Now, as we have the centre and the radius, we can write the equation of the circle.

We have (h, k) = (4,3) and r= 3

So,

(x – 4)² + (y – 3)² = 3²

(x – 4)² + (y – 3)² = 9

Thus, now we have the standard form of equation of circle S.

Example 2) Now for second example:

Now here we have to write the standard form of equation for the circle C with centre at (-4, -1)

So, the information which we need to solve this question is the location of the centre of the circle and the length of the radius of the circle.

The location is known with coordinates as C= (-4, -1)

And to find radius we have to start with c move till the circumference of circle counting the units that is 5 units. So, radius r = 5.

Now let’s move to the equation

(Here double negative becomes positive as per rule)

(x + 4))² + (y + 1)² = 25

And thus, we get the final standard equation of a circle C.

Example 3) Now let’s take one more example with centre as origin

Here we have to write the standard form of equation for Circle J with centre of the origin and diameter 18.

So let start writing the standard form of equation of a circle.

\((x - h)^2 + (y - k)^2 = r^2 \)

Here we need the centre and the radius.

In this case the centre of the circle is at the origin so (h, k) = (0,0)

And radius r = diameter /2

= 18/2

= 9

Let start putting the values in x, y and r position

And we will get

(x – 0)² + (y – 0)² = 9²

(x)² + (y)² = 81

So, the standard equation of a circle with origin at centre is

(x)² + (y)² = 81

Now let’s do one more different example for the standard Equation of a circle.

Example 4) Here we have to identify the centre and the radius of the circle P with a standard graphic equation.

(x – 7)² + (y)² = 196

First try to find the Center and the radius.

Here Center (h= 7, k= 0)

Radius = \(\sqrt{196}\)

= 14

So, radius = 14

Also read: Trapezoid Formula

Examples

[Click Here for Sample Questions]

Q) Find the standard equation of the Circle with centre at (h=2, k=4) and radius = 2

A) Put everything in the formula, and we will get

(x – h)² + (y – k)² = r²

(x – 2)² + (y – 4)² = 2²

(x – 2)² + (y – 4)² = 4

Q) Find the equation of the circle with centre (-2, 3) and radius 5

A) after putting values in the equation, we will get

(x – h)² + (y – k)² = r²

(x – (- 2))² + (y – 3)² = 5²

(x + 2))² + (y – 3)² = 25

Q) Find the centre and radius of the circle

A) The given equation is

(x² + 8x) + (y² + 10y) = 8

After completing the squares within the parenthesis, we will get

(x² + 8x + 16 + y² + 10y + 25 = 8 + 16 + 25

= (x + 4)² + (y + 5)² = 49

= {x – (- 4)}² + {y – (- 5)}² = 49

= {x – (- 4)}² + {y – (- 5)}² = 72

So, the given circle has a centre (-4, -5) and radius 7.

Q) Find the equation of the circle which passes through the point (3, -3) and (2,4) and whose centre lies on the line x + y = 2.

Let the equation of the circle be

\((x - h)^2 + (y - k)^2 = r^2 \)

Since the circle passes through (2,-2) and (3,4), we have

(2 – h)² + ((- 2) – k)² = r²……….(1)

(3 – h)² + (4 – k)² = r²…………..(2)

The centre, as we see, since lies on the line

x + y = 2

So, we have h + k = 2---------------------(3)

Solving equation 1, 2 and 3 we will get

H=0.7, k= 1.3, r²=12.58

Thus, the standard equation of the circle will be

(x – 0.7)² + (y – 1.3)² = 12.58

Now, as we have done many examples, I hope it is clear what is the standard form of equation of a circle and how to solve different problems related to this topic.

Also read: Tan2x Formula

Important Topics for JEE Main

As per JEE Main 2024 Session 1, important topics included in the chapter Standard equation of a circle are as follows:

Tangent of the Circle
Distance Formula

Some memory based important questions asked in JEE Main 2024 Session 1 include:

The value of k for (2k, 3k), (0, 0), (1,0) and (0,1) to be on the circle is:
If the line x + y = 0 is tangent to the circle (x - λ) ² + (y - β)² = 50, then (x + β)² = ?

Important Questions from Standard Equation of a Circle

Ques. What are the parts of a circle? (2 Marks)

Ans.

A circle is a geometric figure which has a centre point and a radius. And when we join all the points in a 360 degree around that centre point, maintaining the same radius, we will get a geometric figure called Circle.
A Radius is the distance between the centre and any point on the circumference of a circle.
A Diameter is a line that crosses the centre of a circle and touches two points on the circumference and hence divides the circle into two equal halves called semi-circles.
The longest line that touches any two points on the circle is called the diameter. Basically, A Chord is that line on a circle that touches any two points on it.
The portion on the circumference of any circle is known as the Arc of the Circle.
A Segment is the region in a circle which is enclosed with the chord and the arc.
A Sector is the part that is formed by enclosing the arc and two radii of a circle.

Ques. There are two parts of a circle, Tangents and Secants. What are these? Explain in short. (4 Marks)

Ans.

Tangent can be defined as the following- It is a line that touches the circle at only one point. The Tangents that are drawn to the circle are sure to be perpendicular to the radii of the circle at the point of contact of the tangent to the circle. Tangent can always be drawn on the outer side of the circle and not from inside

Secant is a line that passes through the circle and touches the circle at two different point.

Ques. What is the equation of a circle? (2 Marks)

Ans. The equation of a circle is written as

\((x - h)^2 + (y - k)^2 = r^2 \)

Where (h, k) is centre and R is radius.

Ques. What is the equation of a circle when the centre is at the origin? (2 Marks)

Ans. The coordinates are (h=0, k=0)

(x – 0)² + (y – 0)² = r²

(x)² + (y)² = r²

Ques. What are the formulas for circles? (2 Marks)

Ans. Circumference of the circle is equal 2 (pi) of radius or pi of diameter.

Area of circle is pi * (radius * radius)

Ques. How do we know if the equation is the equation of a circle? (2 Marks)

Ans. If x and y are squared, and the coefficient of x and y are the same the coefficient of (x)² and (y)² are same. Then it is called an equation of a circle.

We can also represent the standard and general expression of a circle in a complex form.The Standard equation of a circle can be also used to find the radius or diameter of the circle.One more fun fact is that any line, passing through the circle, cannot touch the circle at three or more different points.

Also Read:

Isosceles triangle	Scalene Triangle
Types Of Triangles	Properties of Triangle
Differential Equation	Difference between Sequence and Series
First Order Differential Equation	Differentiation and Integration Formula
Eccentricity	Eccentric Formula
Events in Probability	Probability Distribution

Standard equation of a circle: Formulas, Examples and Important Questions

Formulas

Conic Sections Detailed Video Explanation:

Examples

Important Topics for JEE Main

Important Questions from Standard Equation of a Circle

CBSE CLASS XII Related Questions

1.
Using integration, find the area of the region bounded by the curve \( y = x|x| \), the x-axis, and the vertical lines \( x = -2 \) and \( x = 2 \).

2.
Find:

If \[ (3\hat{i}-2\hat{j}+5\hat{k})\times(4\hat{i}+p\hat{j}+q\hat{k})=\vec{0} \] then find the values of \(p\) and \(q\).

3.
Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).

4.
Find:

The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]

5.
Evaluate:
\[ \int_{0}^{1} \frac{x \tan^{-1}x}{(1+x^2)^{3/2}}\,dx \]

6.
A carpenter needs to design a wooden box in the shape of a cuboid such that the sum of its length and breadth is 3 cm more than its height. Twice of its length, thrice of its breadth and its height add up to 10 cm. Its breadth added to 7 times its height is 1 cm less than 3 times its length.

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Standard equation of a circle: Formulas, Examples and Important Questions

Formulas

Conic Sections Detailed Video Explanation:

Examples

Important Topics for JEE Main

Important Questions from Standard Equation of a Circle

CBSE CLASS XII Related Questions

1.Using integration, find the area of the region bounded by the curve \( y = x|x| \), the x-axis, and the vertical lines \( x = -2 \) and \( x = 2 \).

2. Find: If \[ (3\hat{i}-2\hat{j}+5\hat{k})\times(4\hat{i}+p\hat{j}+q\hat{k})=\vec{0} \] then find the values of \(p\) and \(q\).

3.Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).

4. Find: The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]

5.Evaluate:\[ \int_{0}^{1} \frac{x \tan^{-1}x}{(1+x^2)^{3/2}}\,dx \]

6.A carpenter needs to design a wooden box in the shape of a cuboid such that the sum of its length and breadth is 3 cm more than its height. Twice of its length, thrice of its breadth and its height add up to 10 cm. Its breadth added to 7 times its height is 1 cm less than 3 times its length.

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SUBSCRIBE TO OUR NEWS LETTER

1.
Using integration, find the area of the region bounded by the curve \( y = x|x| \), the x-axis, and the vertical lines \( x = -2 \) and \( x = 2 \).

2.
Find:

If \[ (3\hat{i}-2\hat{j}+5\hat{k})\times(4\hat{i}+p\hat{j}+q\hat{k})=\vec{0} \] then find the values of \(p\) and \(q\).

3.
Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).

4.
Find:

The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]

5.
Evaluate:
\[ \int_{0}^{1} \frac{x \tan^{-1}x}{(1+x^2)^{3/2}}\,dx \]

6.
A carpenter needs to design a wooden box in the shape of a cuboid such that the sum of its length and breadth is 3 cm more than its height. Twice of its length, thrice of its breadth and its height add up to 10 cm. Its breadth added to 7 times its height is 1 cm less than 3 times its length.