Coordinate Geometry Circle

A circle is the set of all points in a plane that are equidistant from a fixed point in the plane.

The fixed point is called the centre of the circle & the distance from the centre to a point on the circle is called the radius of the circle.

 

 

 

 

 

 

 

 

 

 

 

 

 

Given \(C(h,k)\) be the centre & r be the radius of circle. Let \(P(x,y)\) be any point on the circle. \(|CP|=r\). By distance formula:

\(\sqrt { { (x-h) }^{ 2 }+{ (y-k) }^{ 2 } } =r\) \({ (x-h) }^{ 2 }+{ (y-k) }^{ 2 }={ r }^{ 2 }\)

Therefore, any point on the circle satisfies

\({ (x-h) }^{ 2 }+{ (y-k) }^{ 2 }={ r }^{ 2 }\)

Problem: Find the equation of the circle with centre (2,3) & radius 5?

Solution: Equation of the circle is

\({ (x-2) }^{ 2 }+{ (y-3) }^{ 2 }={ 5 }^{ 2 }\\ { x }^{ 2 }-4x+4+{ y }^{ 2 }+9-6y=25\\ { x }^{ 2 }+{ y }^{ 2 }-4x-6y-12=0\)

Important formula:

Let generalized equation of a circle be

\({ x }^{ 2 }+{ y }^{ 2 }+2gx+2fy+c=0\)

Centre of circle is at \((-g,-f)\)

Radius of circle =

\(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c }\)

 

Problem: Find the centre & the radius of the circle

\({ x }^{ 2 }+{ y }^{ 2 }-12x+10y+52=0\)

Solution: Let’s compare this equation with generalized equation:

\({ x }^{ 2 }+{ y }^{ 2 }-12x+10y+52=0\)

2g = -12, c = 52, 2f = 10

g = -6, f = 5, c = 52

Centre (-g,-f) = (6,-5)

Radius = \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } =\sqrt { 36+25-52 }\)

Radius = 3 units

 

 

 

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