Diameter Of A Circle: Definition, Formula, Equation, Examples

In geometry, we have already studied two different types of shapes which included 2D shapes and 3D shapes. From our previous study we know that 2D shapes have only length and breadth. When we talk about a geometric shape like a circle, the condition that defines it is “Diameter”. It is a line of symmetry within the circle. The diameter of a circle helps us to get an idea about the radius, behaviour of circle, chord, etc. In this article, we will learn about diameter of circle and its applications.

This Story also Contains

1. Circle
2. How to Find the Diameter of a Circle?
3. Diameter Properties
4. Difference between radius and diameter of a circle
5. Diameter of a Circle Examples

Circle

A circle is a closed, round geometric shape. Technically, a circle is a point moving around a fixed point at a fixed distance away from the point. It can be said that a circle is a closed curve where its outer line is equal in distance from the center. The fixed distance from the point is known as the radius of the circle. Diameter of a circle is double the length of radius of a circle. If we know the radius, we can easily calculate its diameter.

Diameter of a Circle Definition

Diameter of a circle is the longest line segment which passes through the center of the circle, touching two points of the circumference of circle. It is always perpendicular to the parallel chords. The distance from one point on the surface of a circle to the other point on the surface or circumference of the circle is called the diameter. It is always double the radius. We use various notations to represent diameter such as ‘d’, ‘φ’, ‘D’, etc.

Diameter of a Circle Formula

There exist various ways to calculate the diameter of a circle that are discussed as follows:

$
\mathrm{D}=\mathbf{2 R} \text {,( " } \mathrm{R} \text { "= radius ) }
$

Diameter of a circle formula from circumference:

$\begin{aligned} & \mathrm{D}=\frac{C}{\pi} \\ & (\mathrm{C}=\text { circumference }) \\ & \pi=3.14 \text { (constant })\end{aligned}$

How to Find the Diameter of a Circle?

We take the help of following steps to find the diameter of a circle:

• Step 1: First, we identify the values that are given in the question such as radius, area, etc.
• Step 2: Next, we apply the appropriate formula as discussed in the section above.
• Step 3: At last, we simplify the calculation and get the final answer.

Diameter Properties

The properties of diameter are,

• Chord: Diameter is the longest chord in the circle.
• Relation to the Circumference: Diameter is related to circumference as: $C = \pi \times d$
• Relation to radius: Diameter is double of the radius of the circle.It is related as: $d=2r$
• Symmetry: Diameter divides the circle into two equal parts, which makes it the line of symmetry.

Difference between radius and diameter of a circle

Diameter of a Circle Examples

Example 1: $S(x, y)=0$ represents a circle. The equation $S(x, 2)=0$ gives two identical solutions $x=1$ and the equation $S(1, y)=0$ gives two distinct solutions $y=0,2$. Find the equation of the circle.
1) $x^2+y^2+2 x-2 y+1=0$
2) $x^2+y^2-2 x+2 y+1=0$
3) $x^2+y^2-2 x-2 y-1=0$
4) $x^2+y^2-2 x-2 y+1=0$

Solution
Equation of a circle in diametric form -

$\left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)=0$

- wherein

Where $A\left(x_1, y_1\right)$ and $B\left(x_2, y_2\right)$ are the two diametric ends.

$S(x, 2) = 0$ given two identical solutions x = 1.

$\Rightarrow$ line $y = 2$ is a tangent to the circle $S(x, y) = 0$ at the point $(1, 2)$ and $S(1, y) = 0$ gives two distinct solutions $y = 0, 2$

$\Rightarrow$ Line $x = 1$ cut the circle $S(x, y) = 0$ at points $(1, 0)$ and $(1, 2)$

$A(1, 2)$ and $B(1, 0)$ are diametrically opposite points.

$\therefore$ equation of the circle is $(x-1)^2+y(y-2)=0$

$x^2+y^2-2 x-2 y+1=0$
The correct equation is option (4).

Example 2: The intercept on the line $\mathrm{y}=\mathrm{x}$ by the circle $\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}=0$ is $AB$ . The equation of the circle with $AB$ as a diameter is:
1) $x^2+y^2+x+y=0$
2) $x^2+y^2-x-y=0$
3) $x^2+y^2+x-y=0$
4) none of these

Solution
The line $x=y$ intersect the circle $x^2+y^2-2 x=0$ at $(0,0)$ and $(1,1)$
Hence equation of the required circle is $(x-0)(x-1)+(y-0)(y-1)=0$
Hence, the answer is the option (2).

Example 3: One of the diameters of the circle circumscribing the rectangle ABCD is $4 \mathrm{y}=\mathrm{x}+7$. if $A$ and $B$ are $(-3,4),(5,4)$, the area of the rectangle is

1) $16$ sq. units

2) $24$ sq. units

3) $32$ sq. units

4) $48$ sq. units

Solution

Let $\mathrm{O}(\mathrm{h}, \mathrm{k})$ be the centre of the circle

$ \begin{aligned} & \Rightarrow (h + 3)^2 = (h - 5)^2 \Rightarrow h^2 + 6h + 9 = h^2 - 10h + 25 \\ & \Rightarrow 16h = 16 \Rightarrow h = 1 \\ & (1, k) \text{ lies on } 4y = x + 7 \Rightarrow k = 2 \\ & OB = \sqrt{(1 - 5)^2 + (2 - 4)^2} \\ & = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \end{aligned} $
Let E be the mid-point of $AB$ Then $E$ is $(1,4)$

$\Rightarrow \mathrm{OE}=2 \Rightarrow \mathrm{AB}=8$ and $\mathrm{BC}=4$
$\Rightarrow$ Area $=4 \times 8=32$ sq. units.
Hence (3) is the correct answer.
Example 4: The line $x=2 y$ intersects the ellipse $\frac{x^2}{4}+y^2=1$ at the point $P$ and $Q$. The equation of the circle with $P Q$ as the diameter is
1) $\mathrm{x}^2+\mathrm{y}^2=\frac{1}{2}$
2) $x^2+y^2=1$
3) $x^2+y^2=2$
4) $\mathrm{x}^2+\mathrm{y}^2=\frac{5}{2}$

Solution

$\mathrm{x}=2 \mathrm{y}$
and $\frac{x^2}{4}+y^2=1$

On solving.

$ 2y^2 = 1 \Rightarrow y = \pm \frac{1}{\sqrt{2}} \Rightarrow x = \pm \sqrt{2} $

Therefore $\mathrm{P}\left(\sqrt{2}, \frac{1}{\sqrt{2}}\right)$ and $\mathrm{Q}\left(-\sqrt{2},-\frac{1}{\sqrt{2}}\right)$ (say)
Therefore, Circle with PQ as the diameter is

$ \begin{aligned} & (x - \sqrt{2})(x + \sqrt{2}) + \left(y - \frac{1}{\sqrt{2}}\right)\left(y + \frac{1}{\sqrt{2}}\right) = 0 \\ & \Rightarrow x^2 + y^2 = \frac{5}{2} \end{aligned} $
Hence, the answer is the option (4).

Example 5: Let $\mathrm{A}(-2,0), \mathrm{B}(2,0)$ and $\mathrm{C}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ where $\mathrm{y}_1 \in \mathrm{R}^{+}$be the vertices of an equilateral triangle ABC . If D be any point such that $\angle \mathrm{BDC}=\frac{\pi}{2}$, then the equation of the circumcircle of $\triangle \mathrm{BDC}$ is

1) $x^2+y^2-\sqrt{3} x-2 y=0$
2) $x^2+y^2-2 x-2 \sqrt{3} y=0$
3) $x^2+y^2-2 x-\sqrt{3} y+1=0$
4) None of these

Solution
Clearly $\mathrm{x}_1=0, \mathrm{y}_1=2 \sqrt{3}$
For a right-angled triangle, the circumcircle passes through the hypotenuse which is act like diameter of circumcircle.
$\therefore$ equation of circumcircle of $\mathrm{BDC}=\mathrm{x}(\mathrm{x}-2)+\mathrm{y}(\mathrm{y}-2 \sqrt{3})=0$
Hence, the answer is the option (2).

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