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Position of a Point with Respect to a Circle

We will learn how to find the position of a point with respect to a circle.

A point (x1, y1) lies outside, on or inside a circle S = x2 + y2 + 2gx + 2fy + c = 0 according as S1 > = or <0, where S1 = x12 + y12 + 2gx1 + 2fy1 + c.

Let P (x1, y1) be a given point, C (-g , -f) be the centre and a be the radius of the given circle.

We need to find the position of the point P (x1, y1) with respect to the circle S = x2 + y2 + 2gx + 2fy + c = 0.

Now, CP = (x1+g)2+(y1+f)2

Therefore, the point

(i) P lies outside the circle x2 + y2 + 2gx + 2fy + c = 0 if CP > the radius of the circle.

i.e., (x1+g)2+(y1+f)2 > g2+f2c

(x1+g)2+(y1+f)2 > g2 + f2 - c

⇒ x12 + 2gx1 + g2 + y12 + 2fy1 + f2 > g2 + f2 – c

⇒ x12 + y12 + 2gx1 + 2fy1 + c > 0

⇒ S1 > 0, where S1 = x12 + y12 + 2gx1 + 2fy1 + c.

 

(ii) P lies on the circle x2 + y2 + 2gx + 2fy + c = 0 if CP = 0.

i.e., (x1+g)2+(y1+f)2 = g2+f2c

(x1+g)2+(y1+f)2 = g2 + f2 - c

⇒ x12 + 2gx1 + g2 + y12 + 2fy1 + f2 = g2 + f2 – c

⇒ x12 + y12 + 2gx1 + 2fy1 + c = 0

⇒ S1 = 0, where S1 = x12 + y12 + 2gx1 + 2fy1 + c.

 

(iii) P lies inside the circle x2 + y2 + 2gx + 2fy + c = 0 if CP < the radius of the circle.

i.e., (x1+g)2+(y1+f)2 < g2+f2c

(x1+g)2+(y1+f)2 < g2 + f2 - c

⇒ x12 + 2gx1 + g2 + y12 + 2fy1 + f2 < g2 + f2 – c

⇒ x12 + y12 + 2gx1 + 2fy1 + c < 0

⇒ S1 < 0, where S1 = x12 + y12 + 2gx1 + 2fy1 + c.

Again, if the equation of the given circle be (x - h)2 + (y - k)2 = a2 then the coordinates of the centre C (h, k) and the radius of the circle = a

We need to find the position of the point P (x1, y1) with respect to the circle (x - h)2 + (y - k)2= a2.

Therefore, the point

(i) P lies outside the circle (x - h)2 + (y - k)2 = a2 if CP > the radius of the circle

i.e., CP > a

⇒ CP2 > a2

⇒ (x1 - h)2 + (y1 - k)2 > a2


(ii) P lies on the circle (x - h)2 + (y - k)2 = a2 if CP = the radius of the circle

i.e., CP = a

⇒ CP2 = a2

⇒ (x1 - h)2 + (y1 - k)2 = a2


(iii) P lies inside the circle (x - h)2 + (y - k)2 = a2 if CP < the radius of the circle

i.e., CP < a

⇒ CP2 < a2

⇒ (x1 - h)2 + (y1 - k)2 < a2

 

Solved examples to find the position of a point with respect to a given circle:

1. Prove that the point (1, - 1) lies within the circle x2 + y2 - 4x + 6y + 4 = 0, whereas the point (-1, 2) is outside the circle.

Solution:

We have x2 + y2 - 4x + 6y + 4 = 0 ⇒ S = 0, where S = x2 + y2 - 4x + 6y + 4

For the point (1, -1), we have S1 = 12 + (-1)2 - 4 ∙1 + 6 ∙ (- 1) + 4 = 1 + 1 - 4 - 6 + 4 = - 4 < 0

For the point (-1, 2), we have S1 = (- 1 )2 + 22 - 4 ∙ (-1) +  6 ∙ 2 + 4 = 1 + 4 + 4 + 12 + 4 = 25 > 0

Therefore, the point (1, -1) lies inside the circle whereas (-1, 2) lies outside the circle.

 

2. Discuss the position of the points (0, 2) and (- 1, - 3) with respect to the circle x2 + y2 - 4x + 6y + 4 = 0.

Solution:

We have x2 + y2 - 4x + 6y + 4 = 0 ⇒ S = 0 where S = x2 + y2 - 4x + 6y + 4

For the point (0, 2):

Putting x = 0 and y = 2 in the expression x2 + y2 - 4x + 6y + 4 we have,

S1 = 02 + 22 - 4 ∙ 0 + 6 ∙ 2 + 4 = 0 + 4 – 0 + 12 + 4 = 20, which is positive.

Therefore, the point (0, 2) lies within the given circle.

For the point (- 1, - 3):

Putting x = -1 and y = -3 in the expression x2 + y2 - 4x + 6y + 4 we have,

S1 = (- 1)2 + (- 3)2 - 4 ∙ (- 1) + 6 ∙ (- 3) + 4 = 1 + 9 + 4 - 18 + 4 = 18 - 18 = 0.

Therefore, the point (- 1, - 3) lies on the given circle.

 The Circle




11 and 12 Grade Math 

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