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Circle: Definition and Equation of a Circle

Definition, Equation of a Circle, Practice Problems & FAQs

From bubbles to bangles to doughnuts to pizzas to car wheels there are so many things which are circular in shape. No wonder circles rule our world and command so much attention.

A circle is the set of all points which are at the same distance from a given point. But, since in mathematics every curve is represented by an equation, so is the case with circles also. Let's try to dig more into this.

Table of Contents

What is a circle?

A circle can be defined as the locus of a point that moves in a plane such that its distance from a fixed point in that plane is always constant.

The fixed point is known as the center of the circle and the constant distance is known as the radius of the circle.

Fig: Circle with center at and radius r

Equation of a Circle in different forms

  • Centre-radius form/Cartesian form/Standard form

The equation of the circle whose center is (h, kand radius ‘ r'is (x-h)2+(y-k)2=r2

Note:

  • By the equation of a circle, we mean the equation of circumference of the circle.
  • When the center of the circle coincides with the origin, i.e., h=0 and k=0, the equation becomes x2+y2=r2. We call it the simplest form of a circle equation.
  • Diametric form

The equation of a circle whose end-points of a diameter are (x1, y1and (x2, y2is (x-x1)(x-x2)+(y-y1)(y-y2)=0 i.e x2+y2-(x1+x2)x-(y1+y2)y+x1x2+y1y2=0

Observe that the circle’s equation in diametric form is a combination of quadratic in having roots x1,xand quadratic in having roots y1, y2.

Fig:Circle with endpoints of diameter as (x1,y1& (x2,y2)

Practice Problems

Example : The circle passing through the point (-1, 0) and touching the yaxis at ( 0, 2) also passes through the point

(a) -32, 0
(b) -53,2
(c) -32,52
(d) (-4,0)

Solution :

Step 1 :

Let the given circle be Cwith centre (h,2) and radius units.

(x-h)2+(y-2)2=|h|2

⇒ (x-h)2+(y-2)2=h2

Step 2 :

C1  passes through (-1, 0) 

⇒(-1-h)2+(0-2)2=h2

⇒ h= -52

(x+52)2+(y - 2)2=(52)2.

Only the point (-4, 0) satisfies the equation of the obtained circle, and hence it lies on the circle. Therefore, option (d) is the correct answer.

 


Example : What is the image of the circle (x+8)2+(y-12)2=25 in the line mirror 4x+7y+13=0 ?

(a) (x+16)2+(y+2)2=25

(b) (x-16)2+(y-2)2=25 

(c) (x+16)2+(y-2)2=25  

(d) (x-16)2+(y+2)2=25 

Solution :

Step 1 :

Let C: (x+8)2+(y-12)2=25 

Centre of the circle =(-8, 12)

Radius =5 

Image of centre of Cwith respect to the mirror l:4x+7y+13=0 is the centre of circle Cand radius is same as that of C

 


Step 2 :

Let (h,k)be the image of (-8, 12) w.r.t line l:4x+7y+13=0 

(x2 - x1)a= (y2 - y1)b= -2(ax1 + by1 + c)(a2 + b2

Where a,b,are the coefficients of x,and constant respectively and ( x1, y1is the point whose image is to be found out.
(h +8)4= (k- 12)7= -2(4(-8)+ 7(12)+ 13)(42 + 72)

(h +8)4= (k- 12)7= -2 

h=-16 and k=-2 

Centre of C=(-16, -2) 

Radius = 5 units 

Equation of the required circle is (x+16)2+(y+2)2=25. 

Hence, option (a) is the correct answer.

Example : If the abscissa and ordinates of two points and are the roots of the equation

x2+2ax-b2=0 and x2+2px-q2=0, respectively, then find the equation of the circle with PQ as the diameter.

Solution :

Step 1 :

Let x1, x2, and y1, ybe roots of x2+2ax-b2=0 and x2+2px-q2= 0, respectively.

x1+x2=-2a, x1x2=-b2, y1+y2 =-2p, y1y2 =-q2

Step 2 :

Equation of a circle with (x1,y1) and (x2,y2)  as the end points of the diameter is (x-x1)(x-x2)+(y-y1)(y-y2) = 

x2+y2-x(x1+x2)-y(y1+y2)+x1x2+y1y2= 0

x2+y2+2ax+2py-b2-q2=0 

Therefore, the equation of the required circle is x2+y2+2ax+2py-b2-q2=0. 

Example : What is the equation of a circle that is passing through A(1, 0) and B(0, 1) and having the minimum possible radius?

(a)x2 + y2 + x + y = 0 

(b) x2 + y2 + 2x - y = 0 

(c)x2 + y2 - 2x + y = 0

(d)x2 + y2 - x - y = 0 

Solution :

Step 1 :Let the equation of circle be x2+y2+2gx+2fy+c=0 

Since the circle passes through A(1, 0) and B(0, 1) , these points will satisfy its equation.

1+0+2g+0+c=0 ⇒1+2g+c=0            .....(i)                 

and 0+1+0+2f+c=0⇒1+2f+c=0         .....(ii

Solving (iand (iiwe get g=

The equation of circle becomes x2+y2+2gx+2gy+c=0 

Now the radius of the circle, r=g2+f2-c=g2+g2+1+2g             [ From eq (i) ] 

r=2g2+1+2gr2=2g2+2g+1 

Step 2 : If is minimum then ris minimum

d(r2)dg=0 ⇒4g+2=0⇒g=-12  

Also, the double derivative i.e. d2(r2)d g2=4 which is positive. Hence, g=-1is a point of minima

f=g=-12 & c=-1-2g=0  

Hence the equation of the circle is x2+y2-x-y=0. 

Hence, option (d) is the correct answer.

FAQ’s

1.Is a circle a two-dimensional figure?

Yes, a circle is a two dimensional shape.

2.If A,are two fixed points in a plane, then the locus of a point such that

 is a straight line. Is the statement TRUE or FALSE?

False. The locus will be a circle for given conditions and will represent a straight line if K=1. 

3.What is the circumference of a circle?

If we open a circle to form a straight line, then the length of the obtained line is the

circumference of the circle which is given by 2πr,  here ris the radius of a circle.

4.What are congruent circles?

Two circles with the same radii are called congruent circles.

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