# 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 (x$$_{1}$$, y$$_{1}$$) lies outside, on or inside a circle S = x$$^{2}$$ + y$$^{2}$$ + 2gx + 2fy + c = 0 according as S$$_{1}$$ > = or <0, where S$$_{1}$$ = x$$_{1}$$$$^{2}$$ + y$$_{1}$$$$^{2}$$ + 2gx$$_{1}$$ + 2fy$$_{1}$$ + c.

Let P (x$$_{1}$$, y$$_{1}$$) 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 (x$$_{1}$$, y$$_{1}$$) with respect to the circle S = x$$^{2}$$ + y$$^{2}$$ + 2gx + 2fy + c = 0.

Now, CP = $$\mathrm{\sqrt{(x_{1} + g)^{2} + (y_{1} + f)^{2}}}$$

Therefore, the point

(i) P lies outside the circle x$$^{2}$$ + y$$^{2}$$ + 2gx + 2fy + c = 0 if CP > the radius of the circle.

i.e., $$\mathrm{\sqrt{(x_{1} + g)^{2} + (y_{1} + f)^{2}}}$$ > $$\mathrm{\sqrt{g^{2} + f^{2} - c}}$$

⇒ $$\mathrm{(x_{1} + g)^{2} + (y_{1} + f)^{2}}$$ > g$$^{2}$$ + f$$^{2}$$ - c

⇒ x$$_{1}$$$$^{2}$$ + 2gx$$_{1}$$ + g$$^{2}$$ + y$$_{1}$$$$^{2}$$ + 2fy$$_{1}$$ + f$$^{2}$$ > g$$^{2}$$ + f$$^{2}$$ – c

⇒ x$$_{1}$$$$^{2}$$ + y$$_{1}$$$$^{2}$$ + 2gx$$_{1}$$ + 2fy$$_{1}$$ + c > 0

⇒ S$$_{1}$$ > 0, where S$$_{1}$$ = x$$_{1}$$$$^{2}$$ + y$$_{1}$$$$^{2}$$ + 2gx$$_{1}$$ + 2fy$$_{1}$$ + c.

(ii) P lies on the circle x$$^{2}$$ + y$$^{2}$$ + 2gx + 2fy + c = 0 if CP = 0.

i.e., $$\mathrm{\sqrt{(x_{1} + g)^{2} + (y_{1} + f)^{2}}}$$ = $$\mathrm{\sqrt{g^{2} + f^{2} - c}}$$

⇒ $$\mathrm{(x_{1} + g)^{2} + (y_{1} + f)^{2}}$$ = g$$^{2}$$ + f$$^{2}$$ - c

⇒ x$$_{1}$$$$^{2}$$ + 2gx$$_{1}$$ + g$$^{2}$$ + y$$_{1}$$$$^{2}$$ + 2fy$$_{1}$$ + f$$^{2}$$ = g$$^{2}$$ + f$$^{2}$$ – c

⇒ x$$_{1}$$$$^{2}$$ + y$$_{1}$$$$^{2}$$ + 2gx$$_{1}$$ + 2fy$$_{1}$$ + c = 0

⇒ S$$_{1}$$ = 0, where S$$_{1}$$ = x$$_{1}$$$$^{2}$$ + y$$_{1}$$$$^{2}$$ + 2gx$$_{1}$$ + 2fy$$_{1}$$ + c.

(iii) P lies inside the circle x$$^{2}$$ + y$$^{2}$$ + 2gx + 2fy + c = 0 if CP < the radius of the circle.

i.e., $$\mathrm{\sqrt{(x_{1} + g)^{2} + (y_{1} + f)^{2}}}$$ < $$\mathrm{\sqrt{g^{2} + f^{2} - c}}$$

⇒ $$\mathrm{(x_{1} + g)^{2} + (y_{1} + f)^{2}}$$ < g$$^{2}$$ + f$$^{2}$$ - c

⇒ x$$_{1}$$$$^{2}$$ + 2gx$$_{1}$$ + g$$^{2}$$ + y$$_{1}$$$$^{2}$$ + 2fy$$_{1}$$ + f$$^{2}$$ < g$$^{2}$$ + f$$^{2}$$ – c

⇒ x$$_{1}$$$$^{2}$$ + y$$_{1}$$$$^{2}$$ + 2gx$$_{1}$$ + 2fy$$_{1}$$ + c < 0

⇒ S$$_{1}$$ < 0, where S$$_{1}$$ = x$$_{1}$$$$^{2}$$ + y$$_{1}$$$$^{2}$$ + 2gx$$_{1}$$ + 2fy$$_{1}$$ + c.

Again, if the equation of the given circle be (x - h)$$^{2}$$ + (y - k)$$^{2}$$ = a$$^{2}$$ 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 (x$$_{1}$$, y$$_{1}$$) with respect to the circle (x - h)$$^{2}$$ + (y - k)$$^{2}$$= a$$^{2}$$.

Therefore, the point

(i) P lies outside the circle (x - h)$$^{2}$$ + (y - k)$$^{2}$$ = a$$^{2}$$ if CP > the radius of the circle

i.e., CP > a

⇒ CP$$^{2}$$ > a$$^{2}$$

⇒ (x$$_{1}$$ - h)$$^{2}$$ + (y$$_{1}$$ - k)$$^{2}$$ > a$$^{2}$$

(ii) P lies on the circle (x - h)$$^{2}$$ + (y - k)$$^{2}$$ = a$$^{2}$$ if CP = the radius of the circle

i.e., CP = a

⇒ CP$$^{2}$$ = a$$^{2}$$

⇒ (x$$_{1}$$ - h)$$^{2}$$ + (y$$_{1}$$ - k)$$^{2}$$ = a$$^{2}$$

(iii) P lies inside the circle (x - h)$$^{2}$$ + (y - k)$$^{2}$$ = a$$^{2}$$ if CP < the radius of the circle

i.e., CP < a

⇒ CP$$^{2}$$ < a$$^{2}$$

⇒ (x$$_{1}$$ - h)$$^{2}$$ + (y$$_{1}$$ - k)$$^{2}$$ < a$$^{2}$$

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 x$$^{2}$$ + y$$^{2}$$ - 4x + 6y + 4 = 0, whereas the point (-1, 2) is outside the circle.

Solution:

We have x$$^{2}$$ + y$$^{2}$$ - 4x + 6y + 4 = 0 ⇒ S = 0, where S = x$$^{2}$$ + y$$^{2}$$ - 4x + 6y + 4

For the point (1, -1), we have S$$_{1}$$ = 1$$^{2}$$ + (-1)$$^{2}$$ - 4 ∙1 + 6 ∙ (- 1) + 4 = 1 + 1 - 4 - 6 + 4 = - 4 < 0

For the point (-1, 2), we have S$$_{1}$$ = (- 1 )$$^{2}$$ + 2$$^{2}$$ - 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 x$$^{2}$$ + y$$^{2}$$ - 4x + 6y + 4 = 0.

Solution:

We have x$$^{2}$$ + y$$^{2}$$ - 4x + 6y + 4 = 0 ⇒ S = 0 where S = x$$^{2}$$ + y$$^{2}$$ - 4x + 6y + 4

For the point (0, 2):

Putting x = 0 and y = 2 in the expression x$$^{2}$$ + y$$^{2}$$ - 4x + 6y + 4 we have,

S$$_{1}$$ = 0$$^{2}$$ + 2$$^{2}$$ - 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 x$$^{2}$$ + y$$^{2}$$ - 4x + 6y + 4 we have,

S$$_{1}$$ = (- 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

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