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

The diameter of a circle is one of the most fundamental concepts in geometry. It represents the longest distance across the circle, passing through the centre.

Circle

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 the constant distance is called its radius (r)

DIAMETER OF A CIRCLE

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1. Circle
2. DIAMETER OF A CIRCLE
3. Solved Examples Based on the Diameter of a Circle

The locus of the mid-points of a system of parallel chords of a circle is known as the diameter of the circle.

The diameter of a circle always passes through the centre of a circle and perpendicular to the parallel chords

Let the equation of the circle be $x^2+y^2=a^2$ and equation of parallel chord AB is, $\mathrm{y}=\mathrm{mx}+\mathrm{c}$.

Equation of any diameter to the given circle is perpendicular to the given parallel chord is my + x + λ = 0 which passes through the centre (0, 0) of a circle.

m・0 + 0 + λ = 0

λ = 0

Hence, the required equation of the diameter is x + my = 0

The equation of circle, when endpoints A (x₁, y₁) and B(x₂, y₂) of a diameter are given, is

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

Proof:
$P(x, y)$ is any point on the circle
Slope of $A P=\frac{y-y_1}{x-x_1}$
Slope of BP $=\frac{y-y_2}{x-x_2}$

$
\begin{aligned}
& \because \angle A P B=90^{\circ} \\
& \therefore \text { Slope of AP } \times \text { Slope of } B P=-1 \\
& \Rightarrow\left(\frac{y-y_1}{x-x_1}\right) \times\left(\frac{y-y_2}{x-x_2}\right)=-1 \\
& \Rightarrow\left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)=0
\end{aligned}
$

Properties of the Diameter

1. Longest Chord: The diameter is the longest chord in a circle. A chord is a line segment whose endpoints lie on the circle, and the diameter spans the entire width of the circle.

2. Symmetry: The diameter divides the circle into two equal halves. Each half is a semicircle.

3. Bisects the Circle: Any line that passes through the center of the circle and extends to the circumference will be a diameter. The diameter effectively bisects the circle into two equal parts.

4. Relation to Radius: The diameter of a circle is always twice the radius. If r is the radius, then the diameter D is given by: D=2r

Summary: The diameter of a circle is a fundamental geometric concept with numerous applications in mathematics, science, and everyday life. By understanding its definition, properties, and mathematical relationships, one can solve a wide range of problems involving circles.

Recommended Video Based on the Diameter of a Circle

Solved Examples Based on the Diameter of a Circle

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
$
Maths No Difficulty Level 2193495
The intercept on the line $\mathrm{y}=\mathrm{x}_{\text {by the circle }} \mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}=0$ 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(\mathrm{h}+3)^2=(\mathrm{h}-5)^2 \Rightarrow \mathrm{h}^2+6 \mathrm{~h}+9=\mathrm{h}^2-10 \mathrm{~h}+25 \\
& \Rightarrow 16 \mathrm{~h}=16 \Rightarrow \mathrm{h}=1 \\
& (1, k) \text { lies on } 4 \mathrm{y}=\mathrm{x}+7 \Rightarrow \mathrm{k}=2 \\
& \mathrm{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 A B 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 (C) is the correct answer.
Example 4: The line $x=2 y_{\text {intersects the ellipse }} \frac{x^2}{4}+y^2=1$ 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}
$

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

On solving.

$
2 y^2=1, 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).