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JEE Main 2026 Preparation: Question Papers, Solutions, Mock Tests & Strategy Unacademy » JEE Study Material » Mathematics » The Standard Form of Equation of a Circle

The Standard Form of Equation of a Circle

The graph-based approach of a trigonometric function helps correlate the value of angle and its trigonometric factor.

Table of Content
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A circle is defined as the locus of a point moving in a plane, such that its distance from a fixed point in the plane (i.e., the center) is constant for all such points. The equation of a circle is the algebraic way of expressing a circle if its center and length of the radius are given.

To represent the position of a circle in a Cartesian plane, an equation of a circle is required. The standard form of the equation of a circle is heavily used in coordinate geometry. This equation is different from the formulas used to calculate the circumference or area. 

What is a Circle and Equation of a Circle?

  • A circle is defined as the locus of a point moving in a plane, such that its distance from a fixed point in the plane (i.e., the center) is constant for all such points.
  • It is a closed, two-dimensional shape formed by tracing a point moving in a plane such that its distance from a fixed point is constant. The fixed point is the ‘centre of a circle, and the length of any line drawn from the circle’s center to its boundary is called the ‘radius of a circle.
  • The length of the longest line drawn in a circle is the diameter, i.e., a line segment having endpoints as the circle’s boundary while passing through the center.
  • The equation of a circle is the algebraic way of expressing a circle if its centre and length of the radius are given.
  • To represent the position of a circle in a Cartesian plane, an equation of a circle is required. The equation of a circle is heavily used in coordinate geometry.
  • The circle equation is different from the formulas used to calculate the circumference or area of a circle.

Different Forms of the Equation of a Circle

  • The equation of a circle is the algebraic way of representing the circle’s position in a Cartesian plane.
  • A circle is drawn on paper as long as its centre and the length of the radius is known.
  • Once the coordinates of the circle’s centre and the length of its radius are found using the equation of a circle, one can draw a circle on the Cartesian plane.
  • There are several ways to represent the equation of a circle. They are:
  • Standard form
  • General Form
  • Parametric Form
  • Polar Form
  • The two most common ways of representing a circle are – the general form of the equation of a circle and the standard form of the equation of a circle.
  • The general equation of a circle is: x² + y² + 2gx + 2fy + c = 0, where g, f, and c are constants.

Standard Forms of a Circle

There are various standard forms to represent a circle in a plane. Considering (x, y) as an arbitrary point on the circumference of a circle, the center of the circle as (h, k), and the length of the radius as r in a Cartesian plane, the standard forms of a circle are:

  • Equation of a circle with centre (h, k) is: (x – h)² + (y – k)² = r². If the centre of the circle is its origin i.e., centre is (0, 0), the equation of the circle is x² + y² = r².
  • If the circle passes through the origin, then the equation is: x² + y² – 2hx – 2ky = 0. 
  • The equation of a circle touching the X-axis in a plane is: x² + y² — 2hx — 2ry + h² = 0.
  • The equation of a circle touching the Y-axis in a plane is: x² + y² — 2rx — 2ky + k² = 0.
  • The equation of a circle touching both the X and Y-axes in a plane is: x² + y² — 2rx — 2ry + r² = 0.
  • The equation of a circle passing through the origin and the center lying on the X-axis is: x² + y² – 2rx = 0
  • The equation of a circle passing through the origin and the centre lying on the Y-axis is: x² + y² – 2ry = 0
  • The equation of a circle passing through the origin and cutting intercepts a and b on the coordinate axes is: x² + y² — by = 0.
  • The equation of a circle when the coordinates of endpoints of diameter are (x1, y1) and (x2, y2) is: (x — x1) (x — x2) + (y – y1) (y — y2) = 0.

Standard form Equation of a Circle

  • The standard form of equation of a circle gives precise information about the circle’s centre and its radius, making it easier to read the centre and radius at a glance.
  • Considering (x, y) as an arbitrary point on the circumference of a circle, the center of the circle as (h, k), and the length of the radius as r in a Cartesian plane, the standard form of the equation of a circle is (x – h)² + (y – k)² = r².
  • The distance between this arbitrary point P(x, y) and the center C(h, k) is equal to the radius r of the circle. By applying the distance formula between these points, we get | CP | = r, i.e.,

√{ (x – h)² + (y − k)² }=r

  • On squaring both the sides, we get the standard form of the equation of a circle, i.e.

(x – h)² + (y – k)² = r².

Conclusion

This article covers the concepts, formulas, and scientific terms devised to provide a standard form of the equation of a circle. We introduced the concepts about what a circle is and the equation of a circle to provide the foundation for the central topic, that is, the standard form of the equation of a circle. You are also made familiar with the different forms of equations of a circle and the standard form of a circle, along with formulas for better and easy understanding. In the FAQ section, some solved problems are provided for reference.

faq

Frequently asked questions

Get answers to the most common queries related to the IIT JEE Examination Preparation.

What is a Circle?

Ans : Ans. A circle is defined as the locus of a point moving in a plane, such that its distance from a fixed point in the plane ...Read full

Find the equation of the circle with center (–3, 2) and radius 4.

Ans :  Here h = –3, k = 2 and r = 4. We now substitute these values in the standard form of the equation of a ...Read full

Find the centre and the radius of the circle x² + y² + 8x + 10y – 8 = 0

Ans :  The given equation can be rewritten as (x² + 8x) + (y² + 10y) = 8. Now, completing the squares within the ...Read full

Determine the equation for a circle in standard form with a radius of 3 and centered at the point (−3, 4).

Ans :  The standard form of the equation of a circle with radius r, and centered at the point (h, k) is: (x ...Read full

Ans : Ans. A circle is defined as the locus of a point moving in a plane, such that its distance from a fixed point in the plane (i.e., the center) is constant for all such points. It is a closed, two-dimensional shape formed by tracing a point moving in a plane such that its distance from a fixed point is constant. The fixed point is the ‘center of a circle, and any line drawn from the circle’s center to its boundary is called the ‘radius of a circle.

Ans : 

Here h = –3, k = 2 and r = 4.

We now substitute these values in the standard form of the equation of a circle, (x – h)² + (y – k)² = r².

Therefore, the equation of the required circle with centre (-3, 2) and radius 4 is:

(x + 3)² + (y – 2)² = 16.

Ans : 

The given equation can be rewritten as (x² + 8x) + (y² + 10y) = 8.

Now, completing the squares within the parenthesis, we get

(x² + 8x + 16) + (y² + 10y + 25) = 8 + 16 + 25 i.e.,

(x + 4)² + (y + 5)

The given equation can be rewritten as (x² + 8x) + (y² + 10y) = 8.

Now, completing the squares within the parenthesis, we get

(x² + 8x + 16) + (y² + 10y + 25) = 8 + 16 + 25 i.e.,

(x + 4)² + (y + 5)² = 49 i.e.,

{x – (– 4)}² + {y – (–5)}² = 7²

Therefore, for the given circle, the center is (– 4, –5), and the radius is 7.

= 49 i.e.,

{x – (– 4)}² + {y – (–5)}² = 7²

Therefore, for the given circle, the center is (– 4, –5), and the radius is 7.

Ans : 

The standard form of the equation of a circle with radius r, and centered at the point (h, k) is:

(x − h)

The standard form of the equation of a circle with radius r, and centered at the point (h, k) is:

(x − h)² + (y − k)

The standard form of the equation of a circle with radius r, and centered at the point (h, k) is:

(x − h)² + (y − k)² =r².

Here, h = −3, k = 4, and r = 3.

Substituting these values in the standard form of equation of a circle we get:

(x + 3)² + (y – 4)²= 3².

Hence, the required equation is:

(x + 3)² +(y − 4)² =9.

=r2.

Here, h = −3, k = 4, and r = 3.

Substituting these values in the standard form of equation of a circle we get:

(x + 3)² + (y – 4)² = 3².

Hence, the required equation is:

(x + 3)² +(y − 4)² =9.

+ (y − k)² =r².

Here, h = −3, k = 4, and r = 3.

Substituting these values in the standard form of equation of a circle we get:

(x + 3)² + (y – 4)² = 3².

Hence, the required equation is:

(x + 3)² +(y − 4)² =9.

 

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