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 midpoint of the line segment PQ is (\(\frac{x_{1} + x_{2}}{2}\), \(\frac{y_{1} + y_{2}}{2}\)).
Now see that the midpoint 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 coordinates 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 semicircle 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.
● The Circle
11 and 12 Grade Math
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