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




11 and 12 Grade Math 

From Equation of a Circle when the Line Segment Joining Two Given Points is a Diameter to HOME PAGE




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.




Share this page: What’s this?

Recent Articles

  1. Subtraction of 4-Digit Numbers | Subtract Numbers with Four Digit

    Jan 12, 25 03:23 AM

    Properties of Subtraction of 4-Digit Numbers
    We will learn about the subtraction of 4-digit numbers (without borrowing and with borrowing). We know when one number is subtracted from another number the result obtained is called the difference.

    Read More

  2. 3rd Grade Addition Worksheet | 3-Digit Addition | Word Problems | Ans

    Jan 11, 25 01:06 PM

    3rd Grade Addition Games
    In 3th Grade Addition Worksheet we will solve how to addition of 3-digit numbers without regrouping, addition of three 3-digit numbers without regrouping, addition of 3-digit numbers with regrouping…

    Read More

  3. Addition of 4-Digit Numbers | 4-Digit Addition |Adding 4-Digit Numbers

    Jan 11, 25 03:16 AM

    Addition of 4-Digit Numbers
    We will learn about the addition of 4-digit numbers (without carrying and with carrying). We know how to add 2 or 3, 3-digit numbers without carrying or with carrying.

    Read More

  4. Worksheet on Addition of 4-Digit Numbers | 4 Digit Addition Worksheets

    Jan 11, 25 02:48 AM

    Worksheet on Addition of 4-Digit Numbers
    Practice the questions given in the worksheet on addition of 4-digit numbers. Here we will add two 4-digit numbers (without carrying and with carrying) and three 4-digit numbers

    Read More

  5. Word Problems on 4-Digit Numbers |Addition and Subtraction of 4-Digits

    Jan 10, 25 02:49 PM

    Word Problems on 4-Digit Numbers
    We will solve here some of the word problems on addition and subtraction of 4-digit numbers. We will apply the same method while adding and subtracting the word problems. 1. In a village, there are 25…

    Read More