Diametric Form of a Circle

A circle is one of the most fundamental geometric shapes, consisting of all points in a plane that is equidistant from a fixed point called the center of a circle. It is a very basic shape that is constantly used in mathematics. The main applications of the circle are in geometry, engineering for designing circular instruments, physics, and technology. The diametric form of a circle's equation is derived from the endpoints of a diameter.

This Story also Contains

1. Circle
2. Diametric Form of a Circle
3. Solved Examples Based On Diametric Form of Circle
4. Summary

In this article, we will cover the concept of the diametric form of a circle. This concept falls under the broader category of coordinate geometry. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of two questions have been asked on this concept, including two in 2022.

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)

Diametric Form of a Circle

The equation of circle, when endpoints $\mathrm{A}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ and $\mathrm{B}\left(\mathrm{x}_2, \mathrm{y}_2\right)$ 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

$
\begin{aligned}
&P(x, y) \text { is any point on the circle }\\
&\begin{aligned}
& \text { Slope of } \mathrm{AP}=\frac{\mathrm{y}-\mathrm{y}_1}{\mathrm{x}-\mathrm{x}_1} \\
& \text { Slope of } \mathrm{BP}=\frac{\mathrm{y}-\mathrm{y}_2}{\mathrm{x}-\mathrm{x}_2} \\
& \because \angle \mathrm{APB}=90^{\circ} \\
& \therefore \text { Slope of } \mathrm{AP} \times \text { Slope of } \mathrm{BP}=-1 \\
& \Rightarrow\left(\frac{\mathrm{y}-\mathrm{y}_1}{\mathrm{x}-\mathrm{x}_1}\right) \times\left(\frac{\mathrm{y}-\mathrm{y}_2}{\mathrm{x}-\mathrm{x}_2}\right)=-1 \\
& \Rightarrow\left(\mathrm{x}-\mathrm{x}_1\right)\left(\mathrm{x}-\mathrm{x}_2\right)+\left(\mathrm{y}-\mathrm{y}_1\right)\left(\mathrm{y}-\mathrm{y}_2\right)=0
\end{aligned}
\end{aligned}
$

Recommended Video Based on Diametric Form of Circle

Solved Examples Based On Diametric Form of Circle

Example 1: If $y+3 x=0$ is the equation of a chord of the circle, $x^2+y^2-30 x=0$, then the equation of the circle with this chord as diameter is :
Solution:
First, let us find out the endpoints of this chord of the given circle
Given that the chord is $y+3 x=0$;

$
y=-3 x
$
Its endpoints are its points of intersection with the given circle

$
\begin{aligned}
& x^2+y^2-30 x=0 \\
& x^2+9 x^2-30 x=0 \Rightarrow x=0,3
\end{aligned}
$
Using $\mathrm{y}=-3 \mathrm{x}$, when $\mathrm{x}=0, \mathrm{y}=0$, and when $\mathrm{x}=3, \mathrm{y}=-9$
So the endpoints of the chord are $(0,0)$ and $(3,-9)$
Now these points are the end points of our required circle
So using diametric form, the required circle is

$
\begin{aligned}
& (x-3)(x-0)+(y+9)(y-0)=0 \\
& x^2+y^2-3 x+9 y=0
\end{aligned}
$

Example 2: Let the abscissae of the two points P and Q be the roots of $2 x^2-r x+p=0$ and the ordinates of P and Q be the roots of $x^2-s x-q=0$. If the equation of the circle described on PQ as diameter $2\left(x^2+y^2\right)-11 x-14 y-22=0$, then $2 \mathrm{r}+\mathrm{s}-2 \mathrm{q}+\mathrm{p}$ is equal to $\qquad$
Solution:
Let co-ordinate of P and Q are $(\alpha, \beta)$ and $(\gamma, \delta)$ respectively

$
\begin{aligned}
& \alpha+\gamma=\frac{\mathrm{r}}{2}, \alpha \gamma=\frac{\mathrm{p}}{2} \\
& \text { and } \beta+\delta=\mathrm{s}, \beta \delta=-\mathrm{q}
\end{aligned}
$
Equation of a circle with $\mathrm{PQ}{\text { as diameter }}$

$
\begin{aligned}
& (\mathrm{x}-\alpha)(\mathrm{x}-\gamma)+(\mathrm{y}-\beta)(\mathrm{y}-\delta)=0 \\
& \mathrm{x}^2+\mathrm{y}^2-\mathrm{x}(\alpha+\gamma)-\mathrm{y}(\beta+\delta)+\alpha \gamma+\beta \delta=0 \\
& \mathrm{x}^2+\mathrm{y}^2-\mathrm{x}\left(\frac{\mathrm{r}}{2}\right)-\mathrm{y}(\mathrm{S})+\frac{\mathrm{p}}{2}-\mathrm{q}=0---(\mathrm{I})
\end{aligned}
$
$
\begin{aligned}
& S: 2\left(x^2+y^2\right)-11 x-14 y-22=0 \\
& S: x^2+y^2-\frac{11}{2} x-7 y-11=0----(\text { II })
\end{aligned}
$

$\because$ eq(I) & (II)represent same circle

$
\begin{gathered}
\therefore \frac{r}{2}=\frac{11}{2} \\
r=11
\end{gathered}|S=7| \begin{gathered}
\frac{P}{2}-q=-11 \\
P-2 q=-22
\end{gathered}
$
$
\begin{aligned}
\therefore & 2 r+S-2 q+P \\
& 2 \times 11+7-22=7
\end{aligned}
$
Hence, the answer is 7.

Example 3: The intercept on the line $y=x$ by the circle $x^2+y^2-2 x=0$ is $A B$. The equation of the circle with $A B$ 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 $\mathrm{x}=\mathrm{y}_{\text {intersect the circle }} \mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}=0$ 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 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
Solution:

$
x=2 y
$

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

On solving,

$
\begin{aligned}
& \qquad 2 \mathrm{y}^2=1, \mathrm{y}= \pm \frac{1}{\sqrt{2}} \Rightarrow \mathrm{x}= \pm \sqrt{2} \\
& \text { On solving, } \\
& \therefore \mathrm{P}\left(\sqrt{2}, \frac{1}{\sqrt{2}}\right) \text { and } \mathrm{Q}\left(-\sqrt{2},-\frac{1}{\sqrt{2}}\right) \text { (say) }
\end{aligned}
$

$\therefore$ 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}
$

Summary

The circles are foundational shapes with unique properties and applications in various mathematics, science, and engineering fields. Understanding the properties, equations, and applications of circles is essential for solving geometric problems, designing objects, and analyzing natural phenomena.