BRANCH- SC &HUM.

SEM -1ST

SUBJECT-MATHEMATICS

FACULTY- Sushanta madhei

• Equation of Straight Line

The relation between variables x, y satisfy all points on the curve.

The general equation of straight line is as given below:

ax + by + c = 0 {equation of straight line}

where x, y are variables and a,b, c are constants.

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Slope and Y-intercept Form

• Slope and Y-intercept Form
• A straight line having slope m = tanθ where θ is the angle formed by the line with the positive x-axis, and y-intercept as c is given by:

y = mx + c where m is the slope.

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Slope Point Form�

• Slope Point Form

A straight line having slope m = tanθ where θ is the angle formed by the line with the positive x-axis, and passing through a point (x1,y1) is given by: 

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y−y1=m(x−x1)

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Intercept Form

Intercept Form

A straight line having x-intercept as a and y-intercept as b as shown in the figure where point B is on the x-axis (horizontal here) and point A is on the y-axis (vertical here), is given in the intercept form by

x/a + y/b = 1

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CIRCLE

• Definition : Circle is the locus of points which are at a constant distance from a fixed point.
• Note 1 : This constant distance is called as radius of the circle

Note 2 : The fixed point is called the center of the circle.

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• The equation of circle provides an algebraic way to describe a circle, given the center and the length of the radius of circle. The equation of a circle is different from the formulas that are used to calculate the area or the circumference of a circle. This equation is used across many problems of circles in coordinate geometry.
• To represent a circle on the Cartesian plane, we require the equation of the circle. A circle can be drawn on a piece of paper if we know its center and the length of its radius. Similarly, on a Cartesian plane, we can draw a circle if we know the coordinates of the center and its radius. A circle can be represented in many forms:

(i) General form

(ii) Standard form

(iii) Diametric form

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Center Radius Form

• The equation of a circle formula is used for calculating the equation of a circle. We can find the equation of any circle, given the coordinates of the center and the radius of the circle by applying the equation of circle formula. The equation of circle formula is given as, 

(x−h)²+(y−k)²= r²

where,(h,k) is the center of the circle with radius r and (x, y) is an arbitrary point on the circumference of the circle.

• Example 1 : Let point (1,2) is the center of the circle and radius of the circle is equal to 4 cm. Then the equation of this circle will be:

(x-1)²+(y-2)² = 4²

= > (x²−2x+1)+(y²−4y+4) =16

=> X²+y²−2x−4y-11 = 0    

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Equation of a Circle When the Centre is Origin�

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Consider an arbitrary point P(x, y) on the circle. Let ‘a’ be the radius of the circle which is equal to OP.

We know that the distance between the point (x, y) and origin (0,0)can be found using the distance formula which is equal to

√[x²+ y²]= a

Therefore, the equation of a circle, with the center as the origin is,

x²+y²= a²

Where “a” is the radius of the circle.

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• Example 1:

Find the equation of the circle whose centre is (3,5) and the radius is 4 units.

Solution:

Here, the centre of the circle is not an origin.

Therefore, the general equation of the circle is,

(x-3)² + (y-5)²  = 4²

=› x² – 6x + 9 + y² -10y +25 = 16

=› x² +y² -6x -10y + 18 =0

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Example 1:

Consider a circle whose centre is at the origin and radius is equal to 8 units.

Solution:

Given: Centre is (0, 0), radius is 8 units.

We know that the equation of a circle when the centre is origin

x²+ y² = a²

For the given condition, the equation of a circle is given as

x² + y² = 8²

=› x² + y²= 64,

which is the equation of a circle

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General form of a Circle

• The general equation of any type of circle is represented by :

x² + y² + 2gx + 2fy + c = 0, for all values of g, f and c.

Adding g² + f² on both sides of the equation gives,

x² + 2gx + g²+ y² + 2fy + f²= g² + f² − c ………………(1)

Since, (x+g)² = x²+ 2gx + g² and (y+f)² =y² + 2fy + f² substituting the values in equation (1), we have

(x+g)²+ (y+f)² = g² + f²−c …………….(2)

Comparing (2) with (x−h)² + (y−k)² = r², where (h, k) is the center and ‘r’ is the radius of the circle.

h=−g, k=−f and r² = g²+ f²−c

Therefore, x² + y² + 2gx + 2fy + c = 0, represents the circle with centre (−g,−f) and radius equal to r² = g² + f²− c.

* Special Notes

• If g² + f² > c, then the radius of the circle is real.
• If g² + f² = c, then the radius of the circle is zero which tells us that the circle is a point which coincides with the centre. Such type of circle is called a point circle
• g² + f² <c, then the radius of the circle become imaginary. Therefore, it is a circle having a real centre and imaginary radius.

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Example 1 :

Consider an example here to find the center and radius of the circle from the general equation of the circle: x² + y² - 6x - 8y + 9 = 0.

Solution :

The coordinates of the center of the circle can be found as: (-g,-f).

Here g = -6/2 = -3 and f = -8/2 = -4. So, the center is (3,4).

Radius r = √g2+f2−cg2+f2−c 

=›  √(−3)2+(−4)2−9(−3)2+(−4)2−9

  =›  √9+16−99+16−9 =

=›  √16 = 4.

So, radius r = 4.

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Example 3:

Equation of a circle is x²+y²−12x−16y+19=0. Find the centre and radius of the circle.

Solution:

Given equation is of the form x²+ y² + 2gx + 2fy + c = 0,

2g = −12,  2f = −16, c = 19

=› g = −6, =› f = −8

Centre of the circle is (-g , -f) = (6,8)

Radius of the circle = √[(−6)² + (−8)² − 19 ]

= √[100 − 19] 

=  √81 = 9 units.

Therefore, the radius of the circle is  9 units.

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