Circle

Definition of circle :

The locus of points such that, the distance of any point on it from fixed point is always constant is called as circle.

Figure 1

Here, fixed point is called centre and fixed distance is called radius of circle.

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Centre - Radius form of equation of circle :

Figure 2

Consider circle as shown in fig. 2 with centre at C(h,k) and passing through point P(x,y) Join point C and P. So, CP is radius of circle ie. CP = a.

Now, by distance formula, we get 

Squaring on both sides, we get

This is the equation of circle with centre at (h,k) and radius equal to a.

Moreover, if centre of the circle is at origin ie. (0,0) then equation (1) becomes,

 This is the standard equation of circle.

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Ex. 

Find the equation of circle having centre at (3,2) and passing through (2,-1).

Solution :-

Given that, centre of the circle is C(3,2)≡(h,k) and it passes through P(2,-1) as shown in fig.

Figure 3

 

Now, by distance formula, we get

We know that, centre-radius form of equation of circle is,

This is the required equation of circle.

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Diameter form of equation of circle :

Figure 4

 As shown in fig. 4, let,

be two end points of diameter of circle. Let, 

be any point on circle and C be the centre of circle.

We know that, angle inscribed in semicircle is always right angle.

and AP is perpendicular to BP.

Since,  AP is perpendicular to BP.

(Slope of AP) X (Slope of BP) = -1.

 

This is the required equation of circle.

 

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  Ex.

Find the equation of circle, having (-1,2) and (3,-4) as the end points of a diameter.

Solution:

Given that,

as the end points of diameter of circle.

We know that, diameter form of equation of circle is given by,

This is the required equation of circle.

 

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Ex.

Find the equation of the circle whose centre is at (3,-4) and the line 3x - 4y - 5 = 0 cuts the circle at A and B where, 

Solution :

Figure 5

Given that, centre of the circle is C (3,-4) = (h,k) and the line         3x - 4y - 5 = 0 cuts circle at A and B. Also given that, 

Draw CM perpendicular to AB. Therefore, AM = MB = 3.

Now, a perpendicular distance of C(3,-4) from line 3x - 4y - 5 = 0

is given by,

Triangle CMA is right angled triangle. 

By, Pythagorus theorem, we have

We know that, centre - radius form of equation of circle is,

This is the required equation of circle.

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General equation of circle :

We know that, Centre-Radius form of equation of circle is given by,

Comparing it with second degree equation given by,

We get,

Thus, the general equation of circle is given by,

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Ex.

Show that the equation

represents a circle. Also, find radius and centre.

Solution :

Given equation can be written as, 

Dividing both sides by 3, we get

Which is of the form,

Comparing equation (1) and general equation of circle, we get

This shows that, given equation represents a circle.

Also, Centre of circle is (-g,-f) = (-2,-3)

Hence proved.

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Ex.

Find the equation of circle passing through the points (5,7), (6,6), and (2,-2).

Solution :

Figure 6

 Given that, circle passes through 

Let, centre of the circle is

By distance formula, we have

 Since, CP = CQ = CR = Radius = r. We have

Solving equation (1) and (2) simultaneously, we get 

Put the value of k = 3 in equation (2), we get

 

Now,

We know that, centre-radius form of equation of circle is given by,

This is the required equation of circle.

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Ex.

Show that the points (3,-2), (1,0), (-1,-2) and (1,-4) lies on the same circle ie. they are concyclic.

Solution : 

General equation of circle is given by,

Suppose that, (3,-2), (1,0) and (-1,-2) lies on circle given by equation (1)

Now, equation (2) - 3 (equation (3)), we get

 Equation (3) - Equation (4), we get

Put c = 1 in equation (3) we get

                              2g + 2 = 0      ie. g = -1

Put c = 1 and g = -1 in equation (2) we get

                          (-6) - 4f + 14 = 0

                                  - 4f = -8      ie. f = 2

Therefore, equation (1) becomes

Put (1,-4) in equation (7) we get

This shows that, (1,-4) satisfy equation (7) which implies that, point (1,-4) lies on circle given by equation (1).

Thus, (3,-2), (1,0), (-1,-2) and (1,-4) lies on same circle. Hence these points are concyclic.

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Parameter :

Definition:

 If co-ordinates of a point on the curve are expressed as function of another variable then that variable is called parameter of the curve.

Parametric form of equation of circle :

Figure 7

As shown in fig. 7 consider a circle with centre at origin ie. O(0,0)

Let, P(x,y) be any point on circle whose distance from origin is r. 

PN is perpendicular to x-axis which meets axis at N.

Now, △ ONP is right angled.

∴ we have,

Thus, parametric form of equation of circle

Moreover, the parametric form of equation of circle 

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  Tangent to Circle :

A straight line which intersects a circle in coincident point is called as tangent to circle.

The point of intersection is called point of contact.

Show that, the equation of tangent to the circle

at a point

Solution :

Given equation of circle is

Equation of tangent in

parametric form:

 

Equation of tangent to the circle 

at a point P(θ) = (rcosθ,rsinθ)