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Equation of a Circle when the Line Segment Joining Two Given Points is a Diameter

We will learn how to find the equation of the circle for which the line segment joining two given points is a diameter.

the equation of the circle drawn on the straight line joining two given points (x$_{1}$, y$_{1}$) and (x$_{2}$, y$_{2}$) as diameter is (x - x$_{1}$)(x - x$_{2}$) + (y - y$_{1}$)(y - y$_{2}$) = 0

First Method:

Let P (x$_{1}$, y$_{1}$) and Q (x$_{2}$, y$_{2}$) are the two given given points on the circle. We have to find the equation of the circle for which the line segment PQ is a diameter.

Equation of a Circle when the Line Segment Joining Two Given Points is a Diameter

Therefore, the mid-point of the line segment PQ is ($\frac{x_{1} + x_{2}}{2}$, $\frac{y_{1} + y_{2}}{2}$).

Now see that the mid-point of the line segment PQ is the centre of the required circle.

The radius of the required circle

= $\frac{1}{2}$PQ

= $\frac{1}{2}$$\mathrm{\sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}}$

We know that the equation of a circle with centre at (h, k) and radius equal to a, is (x - h)$^{2}$ + (y - k)$^{2}$ = a$^{2}$.

Therefore, the equation of the required circle is

(x - $\frac{x_{1} + x_{2}}{2}$)$^{2}$ + (y - $\frac{y_{1} + y_{2}}{2}$)$^{2}$ = [$\frac{1}{2}$$\mathrm{\sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}}$ ]$^{2}$

⇒ (2x - x$_{1}$ - x$_{2}$)$^{2}$ + (2y - y$_{1}$ - y$_{2}$)$^{2}$ = (x$_{1}$ - x$_{2}$)$^{2}$ + (y$_{1}$ - y$_{2}$)$^{2}$

⇒ (2x - x$_{1}$ - x$_{2}$)$^{2}$ - (x$_{1}$ - x$_{2}$)$^{2}$ + ( 2y - y$_{1}$ - y$_{2}$ )$^{2}$ - (y$_{1}$ - y$_{2}$)$^{2}$ = 0

⇒ (2x - x$_{1}$ - x$_{2}$ + x$_{1}$ - x$_{2}$)(2x - x$_{1}$ - x$_{2}$ - x$_{1}$ + x$_{2}$) + (2y - y$_{1}$ - y$_{2}$ + y$_{1}$ - y$_{2}$)(2y - y$_{1}$ - y$_{2}$ + y$_{2}$) = 0

⇒ (2x - 2x$_{2}$)(2x - 2x$_{1}$) + (2y - 2y$_{2}$)(2y - 2y$_{1}$) = 0

⇒ (x - x$_{2}$)(x - x$_{1}$) + (y - y$_{2}$)(y - y$_{1}$) = 0

⇒ (x - x$_{1}$)(x - x$_{2}$) + (y - y$_{1}$)(y - y$_{2}$) = 0.

Second Method:

equation of a circle when the co-ordinates of end points of a diameter are given

Let the two given points be P (x$_{1}$, y$_{1}$) and Q (x$_{2}$, y$_{2}$). We have to find the equation of the circle for which the line segment PQ is a diameter.

Equation of Circle when the Line Segment Joining Two Given Points is a Diameter

Let M (x, y) be any point on the required circle. Join PM and MQ.

m$_{1}$ = the slope of the straight line PM = $\frac{y - y_{1}}{x - x_{1}}$

m$_{2}$ = the slope of the straight line PQ = $\frac{y - y_{2}}{x - x_{2}}$.

Now, since the angle subtended at the point M in the semi-circle PMQ is a right angle.

Now, PQ is a diameter of the required circle.

Therefore, ∠PMQ = 1 rt. angle i.e., PM is perpendicular to QM

Therefore, $\frac{y - y_{1}}{x - x_{1}}$ × $\frac{y - y_{2}}{x - x_{2}}$ = -1

⇒ (y - y$_{1}$)(y - y$_{2}$) = - (x - x$_{1}$)(x - x$_{2}$)

⇒ (x - x$_{1}$)(x - x$_{2}$) + (y - y$_{1}$)(y - y$_{2}$) = 0.

This is the required equation of the circle having (x$_{1}$, y$_{1}$) and (x$_{2}$, y$_{2}$) as the coordinates of the end points of a diameter.

Note: If the coordinates of the end points of a diameter of a circle given, we can also find the equation of the circle by finding the coordinates of the centre and radius. The centre is the mid-point of the diameter and radius is half of the length of the diameter.

● The Circle

11 and 12 Grade Math

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