EQUATION OF CIRCLE WITH CENTER AND RADIUS

Standard form equation of a circle :

(x - h)² + (y - k)²  =  r²

Here (h, k) stands for center of the circle and "r" stands for radius.

If the center of the circle is at origin, the equation will become

 x ² + y ²  =  r²

Example 1 :

Find the equation of the circle if the center and radius are (2, − 3) and 4 respectively.

Solution :

Since center of the circle is at the point (2, -3), the equation of the circle is

 (x - h)² + (y - k)²  =  r²

Here (h, k) ==>  (2, -3) and r  =  4

 (x - 2)² + (y - 3)²  =  4²

By comparing (x - 2)² and (y - 3)² with the algebraic identity (a - b)², we get

x² - 2(x)(2) + 2² + y² - 2(y)(3) + 3²  =  4²

x² - 4x + 4 + y² - 6y + 9  =  16

x² + y²  - 4x - 6y + 13  =  16

x² + y²  - 4x - 6y + 13 - 16  =  0

x² + y²  - 4x - 6y - 3  =  0

Hence the required equation of the circle is x² + y²  - 4x - 6y - 3  =  0

Example 2 :

Find the equation of the circle with center (-2, 5) and radius 3. Show that it passes through the point (2, 8).

Solution :

Since center of the circle is at the point (2, -3), the equation of the circle is

 (x - h)² + (y - k)²  =  r²

Here (h, k) ==>  (-2 , 5) and r  =  3

 (x - (-2))² + (y - 5)²  =  3²

 (x + 2)² + (y - 5)²  =  3²

By comparing (x + 2)² and (y - 5)² with the algebraic identity (a + b)² and (a - b)², we get

x² - 2(x)(2) + 2² + y² - 2(y)(5) + 5²  =  3²

x² - 4x + 4 + y² - 10y + 25  =  9

x² + y² - 4x - 10y + 29  =  9

x² + y² - 4x - 10y + 29 - 9  =  0

x² + y² - 4x - 10y + 20  =  0

Hence the required equation of the circle is x² + y² - 4x - 10y + 20  =  0

To show that the circle is passing through the point (2, 8), we need to apply this point in the above equation.

x = 2 and y = 8

2² + 8² - 4(2) - 10(8) + 20  =  0

4 + 64 - 8 - 80 + 20  =  0

88 - 88  =  0

0 = 0

Since the point (2, 8) satisfies the above equation, we can say that the point (2, 8) is passing through the circle.

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