FINDING THE EQUATION OF A CIRCLE IN GENERAL FORM

Here we are going to see some example problems to know how to find equation of a circle in general form.

Standard equation of a circle with centre (0, 0) and radius r :

x2 + y2  =  r2

Equation of a circle with centre (h, k) and radius r :

(x - h)2 + (y - k) =  r2

General form of the equation of a circle : 

x2 + y2 + 2gx + 2fy + c  =  0

Example 1 :

Obtain the equation of the circles with radius 5 cm and touching x-axis at the origin in general form.

Solution :

Equation of a circle with centre (h, k) and radius r :

(x - h)2 + (y - k) =  r2

The center lies on y axis. The center point will be at (0, 5) and (0, -5).

Center (0, 5) and r = 5

(x - h)2 + (y - k) =  r2

(x - 0)2 + (y - 5) =  52

x2 + y2 - 10y + 25 - 25  = 0

x2 + y2 - 10y  = 0

Center (0, -5) and r = 5

(x - h)2 + (y - k) =  r2

(x - 0)2 + (y + 5) =  52

x2 + y2 + 10y + 25 - 25  = 0

x2 + y2 + 10y  = 0

So, the required equations are 

x2 + y2 - 10y  =  0 and x2 + y2 + 10y  =  0

Example 2 :

Find the equation of the circle with centre (2, -1) and passing through the point (3, 6) in general form.

Solution :

Equation of a circle with centre (h, k) and radius r :

(x - h)2 + (y - k) =  r2

Centre (h, k)  ==>  (2, -1).

(x - 2)2 + (y + 1) =  r2 -----(1)

The given circle is passing through the point (3, 6). 

Then, substitute 3 for x and 6 for y. 

(3 - 2)2 + (6 + 1) =  r2

12 + 7 =  r2

1 + 49  =  r2

50  =  r2

Then, 

 (1)-----> (x - 2)2 + (y + 1) =  50

x2 - 2(x)(2) + 22 + y2 + 2(y)(1) + 12  =  50

x2 - 4x + 4 + y2 + 2y + 1  =  50

x+ y2 - 4x + 2y + 5  =  50

Subtract 50 from each side. 

x+ y2 - 4x + 2y - 45  =  0

Example 3 :

Find the equation of circles that touch both the axes and pass through (-4, -2) in general form

Solution :

The center point will be at (-r, -r)

By applying the point passes through the circle and center, we get

(x - h)2 + (y - k) =  r2  

(-4 + r)2 + (-2 + r) =  r2  

16 + r2 - 8r + 4 - 4r + r2 - r2  =  0

20 + r2 - 12r  =  0

r2 - 12r + 20  =  0

(r - 10) (r - 2)  =  0

r  =  10 and r  =  2

Equation of a circle center is at (-10, -10) and radius is 10.

(x + 10)2 + (y + 10) =  102  

x2 + 20x + 100 + y2 + 20y + 100 - 100  =   0

x2 + 20x + y2 + 20y + 100  =  0

x2 + y2 + 20x + 20y + 100  =  0

Equation of a circle center is at (-2, -2) and radius is 2.

(x + 2)2 + (y + 2) =  22  

x2 + 4x + 4 + y2 + 4y + 4 - 4  =   0

x2 + 4x + y2 + 4y + 4  =  0

x2 + y2 + 4x + 4y + 4  =  0

Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Problems on Finding Derivative of a Function

    Mar 29, 24 12:11 AM

    Problems on Finding Derivative of a Function

    Read More

  2. How to Solve Age Problems with Ratio

    Mar 28, 24 02:01 AM

    How to Solve Age Problems with Ratio

    Read More

  3. AP Calculus BC Integration of Rational Functions by Partical Fractions

    Mar 26, 24 11:25 PM

    AP Calculus BC Integration of Rational Functions by Partical Fractions (Part - 1)

    Read More