**Inside outside or on the circle :**

Here we are going to see how to determine if a point is inside or outside a circle.

Length of the tangent = √ (x₁² + y₁² + 2gx₁ +2fy₁ + c)

To check if the given point lie on the circle, inside the circle or out side the circle, we use the formula for length of tangent.

(i) If the length is 0.Then we can say, the given point must lie on the circle.

√ (x₁² + y₁² + 2gx₁ +2fy₁ + c) = 0

(ii) If the length is > 0 .Then we can say, the point must lie outside the circle.

√ (x₁² + y₁² + 2gx₁ +2fy₁ + c) > 0

(iii) If the length is < 0 .Then we can say, the point must lie inside the circle.

√ (x₁² + y₁² + 2gx₁ +2fy₁ + c) < 0

**Example 1 :**

Determine whether the points (− 2, 1), (0, 0) and (4, − 3) lie outside, on or inside the circle x^{2} + y^{2} − 5x + 2y − 5 = 0

**Solution :**

(i) First let us check where does the point (-2, 1) lie on the circle x^{2} + y^{2} − 5x + 2y − 5 = 0

PT = √(x₁² + y₁² + 2gx₁ +2fy₁ + c)

here x_{1} = -2 and y_{1} = 1

PT = √((-2)^{2} + 1^{2} − 5(-2) + 2(1) − 5)

PT^{2} = (4 + 1 + 10 + 2 − 5)

PT^{2 }= (17 − 5)

PT^{2 }= 12 > 0

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

(ii) (0, 0)

here x_{1} = 0 and y_{1} = 0

PT = √((0)^{2} + 0^{2} − 5(0) + 2(0) − 5)

PT^{2} = (0 + 0 − 0 + 2 − 5)

PT^{2 }= -3 < 0

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

(ii) (4, -3)

here x_{1} = 4 and y_{1} = -3

PT = √(4^{2} + (-3)^{2} − 5(4) + 2(-3) − 5)

PT^{2} = (16 + 9 − 20 - 6 − 5)

PT^{2 }= 25 - 20 - 6 - 5

= 20 - 31

= -11 < 0

Hence the point (4, -3) lies inside the circle.

- How to find the length of tangent
- How to determine if a point is inside or outside a circle
- Find the equation of the tangent to the circle at the point
- How to find the point of intersection of circle and line
- How to prove if a line is tangent to a circle

After having gone through the stuff given above, we hope that the students would have understood "Inside outside or on the circle".

Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.

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