Loading [MathJax]/jax/output/HTML-CSS/jax.js

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 (x1, y1) lies outside, on or inside a circle S = x2 + y2 + 2gx + 2fy + c = 0 according as S1 > = or <0, where S1 = x12 + y12 + 2gx1 + 2fy1 + c.

Let P (x1, y1) 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 (x1, y1) with respect to the circle S = x2 + y2 + 2gx + 2fy + c = 0.

Now, CP = √(x1+g)2+(y1+f)2

Therefore, the point

(i) P lies outside the circle x2 + y2 + 2gx + 2fy + c = 0 if CP > the radius of the circle.

i.e., √(x1+g)2+(y1+f)2 > √g2+f2βˆ’c

β‡’ (x1+g)2+(y1+f)2 > g2 + f2 - c

β‡’ x12 + 2gx1 + g2 + y12 + 2fy1 + f2 > g2 + f2 – c

β‡’ x12 + y12 + 2gx1 + 2fy1 + c > 0

β‡’ S1 > 0, where S1 = x12 + y12 + 2gx1 + 2fy1 + c.

 

(ii) P lies on the circle x2 + y2 + 2gx + 2fy + c = 0 if CP = 0.

i.e., √(x1+g)2+(y1+f)2 = √g2+f2βˆ’c

β‡’ (x1+g)2+(y1+f)2 = g2 + f2 - c

β‡’ x12 + 2gx1 + g2 + y12 + 2fy1 + f2 = g2 + f2 – c

β‡’ x12 + y12 + 2gx1 + 2fy1 + c = 0

β‡’ S1 = 0, where S1 = x12 + y12 + 2gx1 + 2fy1 + c.

 

(iii) P lies inside the circle x2 + y2 + 2gx + 2fy + c = 0 if CP < the radius of the circle.

i.e., βˆš(x1+g)2+(y1+f)2 < √g2+f2βˆ’c

β‡’ (x1+g)2+(y1+f)2 < g2 + f2 - c

β‡’ x12 + 2gx1 + g2 + y12 + 2fy1 + f2 < g2 + f2 – c

β‡’ x12 + y12 + 2gx1 + 2fy1 + c < 0

β‡’ S1 < 0, where S1 = x12 + y12 + 2gx1 + 2fy1 + c.

Again, if the equation of the given circle be (x - h)2 + (y - k)2 = a2 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 (x1, y1) with respect to the circle (x - h)2 + (y - k)2= a2.

Therefore, the point

(i) P lies outside the circle (x - h)2 + (y - k)2 = a2 if CP > the radius of the circle

i.e., CP > a

β‡’ CP2 > a2

β‡’ (x1 - h)2 + (y1 - k)2 > a2


(ii) P lies on the circle (x - h)2 + (y - k)2 = a2 if CP = the radius of the circle

i.e., CP = a

β‡’ CP2 = a2

β‡’ (x1 - h)2 + (y1 - k)2 = a2


(iii) P lies inside the circle (x - h)2 + (y - k)2 = a2 if CP < the radius of the circle

i.e., CP < a

β‡’ CP2 < a2

β‡’ (x1 - h)2 + (y1 - k)2 < a2

 

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 x2 + y2 - 4x + 6y + 4 = 0, whereas the point (-1, 2) is outside the circle.

Solution:

We have x2 + y2 - 4x + 6y + 4 = 0 β‡’ S = 0, where S = x2 + y2 - 4x + 6y + 4

For the point (1, -1), we have S1 = 12 + (-1)2 - 4 βˆ™1 + 6 βˆ™ (- 1) + 4 = 1 + 1 - 4 - 6 + 4 = - 4 < 0

For the point (-1, 2), we have S1 = (- 1 )2 + 22 - 4 βˆ™ (-1) +  6 βˆ™ 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 x2 + y2 - 4x + 6y + 4 = 0.

Solution:

We have x2 + y2 - 4x + 6y + 4 = 0 β‡’ S = 0 where S = x2 + y2 - 4x + 6y + 4

For the point (0, 2):

Putting x = 0 and y = 2 in the expression x2 + y2 - 4x + 6y + 4 we have,

S1 = 02 + 22 - 4 βˆ™ 0 + 6 βˆ™ 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 x2 + y2 - 4x + 6y + 4 we have,

S1 = (- 1)2 + (- 3)2 - 4 βˆ™ (- 1) + 6 βˆ™ (- 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 

From Position of a Point with Respect to a Circle to HOME PAGE




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.




Share this page: What’s this?

Recent Articles

  1. Formation of Square and Rectangle | Construction of Square & Rectangle

    Jul 15, 25 02:46 AM

    Construction of a Square
    In formation of square and rectangle we will learn how to construct square and rectangle. Construction of a Square: We follow the method given below. Step I: We draw a line segment AB of the required…

    Read More

  2. 5th Grade Quadrilaterals | Square | Rectangle | Parallelogram |Rhombus

    Jul 15, 25 02:01 AM

    Square
    Quadrilaterals are known as four sided polygon.What is a quadrilateral? A closed figure made of our line segments is called a quadrilateral. For example:

    Read More

  3. Formation of Numbers | Smallest and Greatest Number| Number Formation

    Jul 14, 25 01:53 AM

    In formation of numbers we will learn the numbers having different numbers of digits. We know that: (i) Greatest number of one digit = 9,

    Read More

  4. 5th Grade Geometry Practice Test | Angle | Triangle | Circle |Free Ans

    Jul 14, 25 01:53 AM

    Name the Angles
    In 5th grade geometry practice test you will get different types of practice questions on lines, types of angle, triangles, properties of triangles, classification of triangles, construction of triang…

    Read More

  5. 5th Grade Circle Worksheet | Free Worksheet with Answer |Practice Math

    Jul 11, 25 02:14 PM

    Radii of the circRadii, Chords, Diameters, Semi-circles
    In 5th Grade Circle Worksheet you will get different types of questions on parts of a circle, relation between radius and diameter, interior of a circle, exterior of a circle and construction of circl…

    Read More