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.

Point Lies Outside 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.

Point Lies On 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.

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

Point Lies Inside 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

• Definition of Circle
• Equation of a Circle
• General Form of the Equation of a Circle
• General Equation of Second Degree Represents a Circle
• Centre of the Circle Coincides with the Origin
• Circle Passes through the Origin
• Circle Touches x-axis
• Circle Touches y-axis
• Circle Touches both x-axis and y-axis
• Centre of the Circle on x-axis
• Centre of the Circle on y-axis
• Circle Passes through the Origin and Centre Lies on x-axis
• Circle Passes through the Origin and Centre Lies on y-axis
• Equation of a Circle when Line Segment Joining Two Given Points is a Diameter
• Equations of Concentric Circles
  
• Circle Passing Through Three Given Points
• Circle Through the Intersection of Two Circles
• Equation of the Common Chord of Two Circles
• Position of a Point with Respect to a Circle
• Intercepts on the Axes made by a Circle
• Circle Formulae
• Problems on Circle

11 and 12 Grade Math 

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