HOW TO DETERMINE IF A POINT IS INSIDE OR OUTSIDE A CIRCLE

About "How to determine if a point is inside or outside a circle"
How to determine if a point is inside or outside a 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.

How to verify that points lie on a circle ?
(i) If the length is 0.Then we can say, the given point must lie on the circle.

√ (x₁² + y₁² + 2gx₁ +2fy₁ + c) = 0

How to verify that points lie outside the circle ?
(ii) If the length is > 0 .Then we can say, the point must lie outside the circle.

√ (x₁² + y₁² + 2gx₁ +2fy₁ + c) > 0

How to verify that points lie inside the circle ?
(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 :

Show that the point (2, 3) lies inside the circle x^{2} + y^{2} − 6x − 8y + 12 = 0.

Solution :

The length of the tangent PT from P(x_{1} , y_{1} ) to the circle

x^{2} + y^{2} + 2gx +2fy + c = 0 is

PT = x_{1} ^{2 } + y_{1} ^{2} + 2gx_{1} + 2fy_{1} + c

PT = √( 22 + 32 − 6.2 − 8.3 + 12)

= √( 4 + 9 − 12 − 24 + 12)

= − 11 < 0

The point (2, 3) lies inside the circle

Example 2 :

Show that the point (2,-1) lies outside the circle x²+y²- 6x - 8y + 12 = 0

Solution :

Length of the tangent

= √ (x₁²
+ y₁² + 2gx₁ +2fy₁ + c)

Here x₁ = 2 and y₁ = -1

= √ (2)² + (-1)² - 6(2) - 8 (-1) + 12

= √ 4 + 1 - 16 + 8 + 12

= √5 + 8 + 12 + 1 - 8

= √ 26 - 8

= √18 = 3√2 units

The length of tangent is positive. So the given point lies outside the circle.

Example 3 :

Is the point (7, − 11) lie inside or outside the circle x^{2} + y^{2} − 10x = 0 ?

Solution :

To know that where does the given point lie in the circle, we have to find the length of tangent.

Length of the tangent

= √ (x₁² + y₁² + 2gx₁ +2fy₁ + c)

Here x₁ = 7 and y₁ = -11

= √ x ^{2} + y ^{2} − 10x

= √7^{2} + (-11) ^{2} − 10(7)

= √49 + 121 - 70

= √ 170 - 70

= √100 = 10 > 0

Hence the given point (7, -11) outside the circle.

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