# Circles - 1

Introduction

Most of the things that we see around us are circular. Sun, moon on full moon day bangle mery-go around, which you loved so much when you where a child; all happened to circles. Ever wondered how a circle can be represented mathematically; well no!!, then we will tell you in this chapter. Also we will take about tangent normulas, chords which we all have hear about. So let us prove deep in to circles.

(x - h)

x

x

Center

ax

this equation refresent circle when,

a = b,h = 0 , g

(1)

(x - h)

x

(2).

x

Why :-

here

(x - h)

x

x

(x - h)

or, x

(4)

x

Why :-

From fig it is clear that radius will be h .

(x - h)

or, x

Center will be(h, h)and radius will be h. But since center would be in any of the four quadrants its coordinates can be taken as radius h.

the circle are :- (x + r)

X

x intercept

Why :-

let it cut the axis x ie y = 0

in points (x

x

x

Intercelt = x

= [(x

similarily .y intercept =

(6) Circle whose diameter is the line joining two point A (s

(xx- x

Why :-

Angle in a semicircle is a right angle

m

or (x - x

__Definition__: locus of a set of points equidistant from a fixed point Equation ofcircle -(x - h)

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

^{2}h_{2}- 2hx + y^{2}- 2ky + k^{2}- r^{2}= 0x

^{2}+ y^{2}+ 2gx + 2fy + C = 0Center

__Genreal second degree equation-__ax

^{2}+ by^{2}+ 2hxy + 2yx +2fy +C = 0this equation refresent circle when,

a = b,h = 0 , g

^{2}+ f^{2}C__Equation of circle in different forms -__(1)

__Centre(h__:-_{1}K) radius a(x - h)

^{2}+ (y - k)^{2}= a^{2}__standard form (when center is origin)__:-x

^{2}+ y^{2}= a^{2}(2).

__center (h,k)and pass through origin-__x

^{2}+ y^{2}- 2hx - 2ky = 0Why :-

here

(x - h)

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

^{2}+ y^{2}- 2hx - 22ky = 0__Center(h,k) and touches the axis of__x-yx

^{2}+y^{2}- 2hx - 2ky + h^{2}= 0(x - h)

^{2}+ (y - k)^{2}= k^{2}or, x

^{2}+ y^{2}- 2hx - l2ky + h^{2}= 0(4)

__Center(h,k) and touches the axis of y__-x

^{2}+ y^{2}-2hx -2ky + k^{2}= 0Why :-

From fig it is clear that radius will be h .

(x - h)

^{2}+ (y - k)^{2}= h^{2}or, x

^{2}+ y^{2}- 2hx - 2ky +k^{2}= 0__Circle which touches both the axis:-__Center will be(h, h)and radius will be h. But since center would be in any of the four quadrants its coordinates can be taken as radius h.

__Illustration -1-__. Find the equation of circle passing through (-2, 3) and touching both the axes.__Solution__- As the circle toucher both theaxed and lies in 2^{nd}quadrant, lits centre is Where r is the radius , Distance of center from (- 2, 3) = radius .the circle are :- (x + r)

^{2}+ (y - r)^{2}= r^{2}X

^{2}+ y^{2}+ 2agx + 2fy +c = 0 ----equation of circle.x intercept

Why :-

let it cut the axis x ie y = 0

in points (x

_{2},0) and (x_{2}, 0)x

_{1}, x_{2}are the roots of x^{2}+ 2yx + c = 0x

_{1}, x_{2}= - 2y , x_{1}. x_{2}= cIntercelt = x

_{2}- x_{1}= [(x

_{2}+ x_{1})^{2}- 4x_{1}x_{2}]^{1/2}=similarily .y intercept =

(6) Circle whose diameter is the line joining two point A (s

_{1}, y_{1})__and B (x__-_{2}, y_{2})__Diametric form__:-(xx- x

_{1})(x - x_{2}) + (y - y_{1}(y - y_{2}) = 0Why :-

Angle in a semicircle is a right angle

m

_{1}m_{2}= - 1or (x - x

_{1})(x - x_{2}) + (y - y_{1})(y - y_{2}) = 0