Basics of Circle

You might already know that Circles have the least surface area but did you ever think about the geometry which makes it so? Well yes, like straight lines, circles have geometry too and trust me when I say, it is as easy as its linear counterpart. After all, how difficult is it to understand a curve. So, let’s roll around the concepts and understand a Circle.


Introduction to Circles
Position of a Point wrt a Circle
Image of a Circle in a Line

Equation of Circle –

Usually, just after the above heading, you’ll find the line written “The equation of a circle is given by…!”. But let’s not do that. We will derive it from scratch.

Let us take a circle of radius ‘r’ and put it on the coordinate plane. Take the centre at O(h,k) and then a point A on the circumference and assume its coordinates to be (x,y) and join OA. Now, we will assume a point B inside the given circle such that OB is ⊥ to AB which gives us a right angled triangle AOB with right angle at B. This makes OA, hypotenuse.

Circle Basics

From Pythagoras Theorem, we can say that

\( OB^2 + AB^2 = OA^2 \)

Here, OA = r; OB = difference in the x-coordinate = x – h; AB = difference in the y-coordinate = y – k

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

and this is the equation of a circle. Now if you’re thinking that this is just at one point, then stop yourself right there. You can take a similar point on any part of the circumference and still you’ll get the same result.

So, this equation is more like a collection of all such points on the circumference and is, hence, the equation of a circle.

The centre of this circle is at (h, k) and if you move it to the origin then the equation will become

 \( x^2 + y^2 = r^2 \)

Equation of circle in parametric form –

Parametric Equation of circle with centre \((h,k)\) and radius R is given by

\( x=h+R \cos{\theta}\) & \( y=k+R \sin{\theta} \)

where θ is the parameter.

Solved Examples for You

Q. 1 Convert the equation \( x^2 + y^2 – 4x + 6y – 12 = 0  \) into standard form and hence find its centre.

Sol – Here we have \( x^2 + y^2 – 4x + 6y – 12 = 0 \)

⇒ \( x^2 – 4x + y^2 + 6y – 12 = 0 \)

Add & subtract 4 & 9 to get perfect squares,

⇒  \( x^2 – 4x + 4 + y^2 + 6y + 9 – 25 = 0 \)

⇒  \( (x – 2)^2 + (y + 3)^2 = 25 \)

⇒  \( (x – 2)^2 + (y – (-3))^2 = 5^2 \)

which is the standard form of equation of circle. Also, the centre is at (2, -3).

Q.2: If we have a circle of radius 20 cm with its centre at the origin, the circle can be described by the pair of equations?

Sol – We know that for parametric form of equation of circle,

\( x=h+R \cos{\theta}\) & \( y=k+R \sin{\theta} \)

Here, since the centre is at (0, 0), so, h = k = 0 and it is already mentioned that radius is 20 cm. Thus, \(x=20 \cos{t}\) & \(y=20 \sin{t}\) are the required pair of equations.
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