# 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.

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.

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

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