HOW TO CHECK IF A GIVEN POINT LIES INSIDE OR OUTSIDE A CIRCLE

Example 1 :

Determine whether the points (-2, 1),(0, 0) and (-4, -3) lie outside, on or inside the circle x² + y² − 5x + 2y − 5 = 0 .

Solution :

Equation of the circle :

x² + y² − 5x + 2y − 5

Equation of the circle at the point (x₁, y₁) :

x₁² + y₁² − 5x₁ + 2y₁ − 5

at (-2, 1)

  =  (-2)² + 1² − 5(-2) + 2(1) − 5

  =  5 + 10 + 2 - 5

  =  12 > 0

So, the point (-2, 1) lies outside the circle.

at (0, 0)

  =  0² + 0² − 5(0) + 2(0) − 5

  =  - 5 < 0

So, the point (0, 0) lies inside the circle.

at (-4, -3)

  =  (-4)² + (-3)² − 5(-4) + 2(-3) − 5

  =  16 + 9 + 20  - 6 - 5

  =  34 > 0

So, the point (-4, -3) lies outside the circle.

Finding Center and Radius of a Circle

Example 2 :

x² + (y + 2)² = 25

Solution :

x² + (y + 2)² = 25

(x - 0)² + (y + 2)² = 5²

(x - h)² + (y - k)² = r²

center (h, k)  =  (0, -2)

radius (r)  =  6

Example 3 :

x² + y² + 6x - 4y + 4 = 0

Solution :

x² + y² + 6x - 4y + 4 = 0

x² + y² + 2gx + 2fy + c = 0

2g = 6 ----> g = 3 ----> -g = -3

2f = -4 ----> f = -2 ----> -f = 2

c = 4

center (-g, -f)  =  (-3, 2)

radius (r)  =  √[g² + f² - c]

=  √[3² + (-2)² - 4]

=  √(9 + 4 - 4)

=  √9

=  3 units

Example 4 :

x² + y² - x + 2y - 3 = 0

Solution :

x² + y² - x + 2y - 3 = 0

x² + y² + 2gx + 2fy + c = 0

2g = -1 ----> g = -1/2 ----> -g = 1/2

2f = 2 ----> f = 1 ----> -f = -1

c = -3

center (-g, -f)  =  (1/2, -1)

radius (r)  =  √[g² + f² - c]

=  √[(1/2)² + (-1)² - (-3)]

=  √(1/4 + 1 + 3)

=  √(1/4 + 4)

=  √(17/2) units

Example 5 :

2x² + 2y² - 6x + 4y + 2  =  0

Solution :

2x² + 2y² - 6x + 4y + 2  =  0

Divide both sides by 2.

x² + y² - 3x + 2y + 1  =  0

x² + y² + 2gx + 2fy + c = 0

2g = -3 ----> g = -3/2 ----> -g = 3/2

2f = 2 ----> f = 1 ----> -f = -1

c = 1

center (-g, -f)  =  (3/2, -1)

radius (r)  =  √[g² + f² - c]

=  √[(-3/2)² + 1² - 1]

=  √(9/4 + 1 - 1)

=  √(9/4)

=  3/2 units

Example 6 :

If the equation 3x² + (3 − p)xy + qy² - 2px = 8pq represents a circle, find p and q . Also determine the center and radius of the circle.

Solution :

The equation x² + y² + 2gx + 2fy + c = 0 is a second degree equation in x and y possessing the following characteristics:

(i) It is a second degree equation in x and y ,

(ii) coefficient of x² = coefficient of y² ≠ 0,

(iii) coefficient of xy = 0 .

3x² + (3 − p) xy + qy² − 2 px = 8pq 

According to (iii), coefficient of x y  = 0

3 - p  =  0

p  =  3

According to (ii), coefficient of x² = coefficient of y² ≠ 0

coefficient of x² = 3 = coefficient of y² 

q = 3

By applying the values of p and q in the given equation, we get

3x² + 3y² − 6x = 72 

x² + y² − 2x = 24 

Center of the circle (-g, -f)

g = -1, f = 0 and c = -24

Center of the circle  =  (1, 0)

Radius of the circle  =  √g² + f² - c

  =  √(-1)² + 0² + 24

  =  √25  =  5

Hence the center and radius of the circle are (1, 0) and 5 respectively.

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