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
11 and 12 Grade Math
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