mathematics. circle - 1 session session objectives
TRANSCRIPT
Mathematics
Circle - 1
Session
Session Objectives
Session Objective
1. Definition -Locus
2. Different forms of circle
-Standard form-End-point diameter form-General form-Parametric form
3. Condition for second degree equation to represent a circle
radius
Definition-Locus
Definition: locus of a point in a plane which is at a fixed given distance from a fixed given point in the plane is circle
.R
C(xc,yc)
x
y P(x,y)C(xc,yc) center; Rradius
P(x,y) Any point on the circle
center
O(0,0)
Definition-Locus
By distance formula
(x-xc)2+(y-yc)2=R2
R
C(xc,yc)
P(x,y)
Locus of any point on the circle
PC2=
This is also called Standard form
.C(xc,yc)
P(x,y)
.C(h,k)
P(x,y)
. .
.
Standard form
Standard form of the circle center at C(xc,yc) and radius R
(x-xc)2+(y-yc)2=R2
x
yWhen origin coincides with center of circle
Corresponding equation is called center at origin form C(0,0)
P(x,y)R
Special Cases:
x2+y2=r2
Case1
Illustrative Problem
Find the equation of the circle having center at (3,4) and passes through (6,7)
Find radius
Radius=distance of points (3,4) & (6,7)=18
Use Standard form
Ans: (x-3)2+(y-4)2=18
Solution:
Illustrative Problem
Find the equation of the circle having x-y=0 and 3x-4y+1=0 as diameters and passes through (2,3)
Solutionx-y=0
3x-4y+1=0Point of intersection of x-y=0 and 3x-4y+1=0
x-y=03x-4y+1=0
By Solving, we get Center(1,1)
Center Of Circle
P(2,3)
O(1,1)
r2=PO2=(2-1)2+(3-1)2=5
Equation of circle: (x-1)2+(y-1)2=5
r
.
A(x1,y1)
B(x2,y2)
y
x
End-Point Diameter Form
Equation of the circle whose end-points of one diameter are given as A(x1,y1) and B(x2,y2)
P(x,y)
Slope of AP=(y-y1)/(x-x1)
Slope of BP=(y-y2)/(x-x2)
(Slope of AP) x (slope of BP)=-1
APBP
Any points on the circle
O(0,0)
End-Point Diameter Form
(y-y1)(y-y2)= - (x-x1)(x-x2)
(x-x1)(x-x2)+(y-y1)(y-y2)=0
.
A(x1,y1)
B(x2,y2)
y
x
P(x,y)
Any points on the circle
End-point diameter form of circle
1 2
1 2
y y y y. 1
x x x x
O(0,0)
Illustrative Problem
Find the centre and radius of the circle represented by
(x-x1)(x-x2)+(y-y1)(y-y2)=0
Two extremities of diameter P(x1,y1) and Q(x2,y2)
Centre is the mid-point of PQ
Radius=CP=CQ
1 2 1 2x x y yCentre C( , )
2 2
.C
P(x1,y1)
Q(x2,y2)
Solution:
2 21 2 1 2PQ = (x x ) (y y )
2 21 2 1 2
1Radius (x x ) (y y )
2
Illustrative Problem
Find the equation of circle of diameter 5 units, passes through (4,0) and has 3x+4y-12=0 as equation of diameter
Solution:Observe:(4,0) lies on 3x+4y-12=0
3x+4y-1
2=0 Equation of the line isx=4+r cos y= 0+r sin Parametric form of line
tan =slope of line=(-3/4)
cos =(-4/5) and sin =(3/5)
(4,0)
and r 5 r 5
Solution Cont.
Let the other end of diameter be (x1,y1)
r= 5 and cos =(-4/5) and sin =(3/5)
x1=4+5 (-4/5)=0 y1=0+ 5 (3/5)=3
or x1=4-5 (-4/5) =8 y1= 0-5 (3/5) =-3
Other end of the diameter is either(0,3) or (8,-3)
for r=5
for r=-5
3x+4y-12=0
(4,0)
(0,3)
(8,-3)(x-0)(x-4)+(y-3)(y-0)=0
or x(x-4)+y(y-3)=0
(x-8)(x-4)+(y+3)(y-0)=0
x2+y2-4x-3y=0
x2+y2-12x+3y+32=0
Solution Cont.
General Form
Standard form:
(x-xc)2+(y-yc)2=R2 ………….(1)
x2+y2-2xcx-2ycy+ xc2+ yc
2- R2=0
In general form it is written as:
x2+y2+2gx+2fy+ c=0 , where g, f, c R
Comparing with (1)
g=-xc,
f=-yc
c= xc2+ yc
2- R2 c-fgR 22
General Form
x2+y2+2gx+2fy+ c=0 represents circle having
center at (-g,-f)
c-fg 22
general form of circle
1. g2+f2-c>0 real circle
2. g2+f2-c=0
3. g2+f2-c<0
(R>0)
?Point circle (R=0)
?Imaginary circle (R<0)
and radius=
Illustrative Problem
Find the centre & radius of the circle 2x2+2y2-6x+6y-5=0
Make co-eff of x2 & y2 as 1
x2+y2-3x+3y-(5/2)=0
g=-(3/2); f=3/2;c=-5/2 Centre(3/2,-3/2)
Centre (3,-3)?
Solution:
2 2Radius (3 / 2) ( 3 / 2) (5 / 2)
7
Note:- center is (-g,-f) when co-eff of x2 & y2 is 1
Illustrative Problem
Show that the four points (1,0), (2,-7),(8,1) and (9,-6) are concylic. i.e,lie on the same circle.
Solution:Let x2+y2+2gx+2fy+c=0 be the equation of the circle.
As it passes through (1,0)
12+02+2g.1+2f.0+c=0 2g+c=-1------(i)
As it passes through (2,-7)22+(-7)2+2g.2+2f.(-7)+c=0 4g-14f+c=-53------(ii)
As it passes through (8,1)
82+12+2g.8+2f.1+c=0 16g+2f+c=-65------(iii)
Solution Cont.
2g+c=-1------(i)4g-14f+c=-53------(ii)
16g+2f+c=-65------(iii)
(ii)-(i) 2g-14f=-52 (ii)-(iii) -12g-16f=12
or, g-7f = -26 or, 3g+4f = -3By Solvingg=-5 ,f=3 c=9
Equation of the circle is: x2+y2-10x+6y+9=0
For (9,-6) : L.H.S= 92+(-6)2-10.9+6.(-6)+9 = 0
Four points are Concylic
Condition For 2nd Degree Equn to Represent Circle
General form of 2nd degree equn:- ax2+by2+2gx+2fy+2hxy+c=0---(1) where a,b,g,f,h,c R
It represents curves of different kind
AS : 1. for a=b=h=0 (and g ,f 0)
2gx+2fy+c=0
equn (1) be straight line
General equation of the circle : x2+y2+2gx+2fy+c=0-------------(ii)
So general 2nd degree equation will represent circle iff
1. Co-eff of x2=Co-eff y2 0
2. Co-eff of xy=0
ax2+by2+2gx+2fy+2hxy+c=0---(i)
General 2nd degree equation:
Condition For 2nd Degree Equn to Represent Circle
a = b 0
h=0
By Comparing (i) and (ii)
Illustrative ProblemGiven that ax2+by2+2gx+2fy+c=0 represents circle. Then find the radius of the circle?
Solution:
As ax2+by2+2gx+2fy+c=0---(1) represents a circle
a=b 0
Equation (1) can be written as 2 2 2g 2fx +y + x+ y+c=0
a a
2 2g f
Radius= ca a
2 2 2g f a ca
2 2 2g f b cb
Illustrative problemFind the equation of the circle passes through the point of intersection of lines x+y=6, 2x+y=4 and x+2y=5
Solution: Let L1 x+y-6=0L2 2x+y-4=0L3 x+2y-5=0
L1=0L 2=0 L
3=0
p1 p2
p3
L1L2+ L1L3+ L3L2=0
What does it represent?
Solution Cont.
L1=0L 2=0 L
3=0
p1 p2
p3
f(x,y)=L1L2+ L1L3+ L3L2
f(x,y)= (x+y-6)(2x+y-4)+(x+y-6)(x+2y-5) +(x+2y-5)(2x+y-4)
(1)
(x1,y1) (x2,y2)
(x3,y3)
As P1(x1,y1) lies on L1=0 and L2=0 x1+y1-6=0 and 2x1+y1-4=0
By putting P1(x1,y1) in (1)
Solution Cont.
L1=0L 2=0 L
3=0
p1 p2
p3
(x1,y1) (x2,y2)
(x3,y3)
f(x1,y1)= (x1+y1-6)(2x1+y1-4)+(x1+y1-6)(x1+2y1-5) +(x1+2y1-5)(2x1+y1-4)
x1+y1-6=0 and 2x1+y1-4=0
= 0x0+ x0x(x1+2y1-5) +(x1+2y1-5)x0
P1 lies on f(x,y)
Similarly it can be proved that P2 and P3 also lies on f(x,y).
= 0
Solution Cont.
P1,P2,P3 lies on f(x,y)=0 points of intersections of L1=0, L2=0 and L3=0 lies on curve f(x,y)=0
L1L2+ L1L3+ L3L2=0
Represents curve, which passes through the point of intersection of L1=0, L2=0 and L3=0
L1=0L 2=0 L
3=0
p1 p2
p3
(x1,y1) (x2,y2)
(x3,y3)
Solution Cont.
L1L2+ L1L3+ L3L2=0
(x+y-6)(2x+y-4) +(x+y-6) (x+2y-5) +(x+2y-5)(2x+y-4)=0
Will represent circle
1.Co-efficient of x2=co-efficient of y2
2++2=1+2+2-------(1)
2.Co-efficient of xy=0
3+3+5=0------------(2)
From (1) and (2): =1 , = -6/5
Solution Cont.
L1L2+ L1L3+ L3L2=0
For =1 and =-6/5 the curve represents circle
Equation of circle is
(x+y-6)(2x+y-4)+(x+y-6)(x+2y-5) -(6/5)(x+2y-5)(2x+y-4)=0
x2+y2 -17x-19y+50 = 0 is the equation of the circle
Parametric Form
For the circle (x-xc)2+(y-yc)2=R2
.C(xc,yc)
P(x,y)
R
xc (R cos)
yc
(R sin)
point P(x,y) on circle represented as:
x= xc + R cos
xc+Rcos
yc+Rsin
collectively represents all the points on circle for different values of : 0 360
w.r.t x-axis
y= yc + R sin
=x
=y
.
P(x,y)
R
Parametric Form
Parametric form
c cx x y y
Rcosθ sinθ
(x-xc)2+(y-yc)2= R2
Special case:
For the circle x2+y2=R2
Parametric form:-
x y= = R
cosθ sinθ
(x = R cosθ; y = R sinθ)
x
y
R cos
R sin
Illustrative Problem
Find the parametric equation of the circle x2+y2-4x-2y+1=0
Find the center(h,k) & radius(r)
h=2, k=1, r=2
Put in the parametric form
x=(2+2cos); y=(1+2 sin )x - 2 y - 1
= = 2cosθ sinθ
Solution:
Class Exercise
Class Exercise -1
Find the equation of the circle whose diameters are 2x–3y+12=0 and x+4y–5=0 and area is 154Solution:
2x-3y+12=0
x+4y-5=0
Point of intersection of x+4y-5=0 and 2x-3y+12=0Center Of CircleO
(-3,2)
2x-3y+12=0x+4y-5=0
By Solving, we get Center(-3,2)
Let the radius of circle is r.
2 222r 154 r 49
7Equation of circle is (x+3)2+(y–2)2=49
Class Exercise - 2
Find the parametric form of the circle: x2+y2+px+py=0.
Solution:x2+y2+px+py=0.
2 2 2p p px y
2 2 2
p px cos
2 2p p
y sin2 2
p por x cos –
22p p
y sin –22
Parametric Form of Equation
Class Exercise - 3
Find the equation of the circle passing through the vertices of the triangle whose sides are along x+y=2, 3x–4y=6 and x–y=0. Find its center and radius.
Solution:Let L1 x+y-2=0 L2 3x-4y-6=0 L3 x-y=0
Equation of circle passes through point of intersection of L1, L2 and L3 is 1 2 2 3 3 1L L L L L L 0
x y – 2 3x – 4y – 6 3x – 4y – 6 x – y
x – y x y – 2 0 (1)
Solution Cont.
x y – 2 3x – 4y – 6 3x – 4y – 6
x – y x – y x y – 2 0.....(1)
Equation (i) will represent circle if
(a) co-efficient of x2 = co-efficient of y2
(b) co-efficient of xy = 0
i.e. 3 3 –4 4 – – 2 – 7 0....(2)
1
i.e. – 1 – 7 0 – ....(3)7
from (2): =-25/7
Solution Cont.
x y – 2 3x – 4y – 6
3x – 4y – 6 x – y
x – y x y – 2 0.....(1)
1 25
and7 7
Equation of circle is:
(x+y–2)(3x–4y–6)–(1/7)(3x–4y–6)(x–y) –(25/7)(x–y)(x+y–2)=0
2 2or, x y 4x 14y 12 0
x2+y2–4x+14y–12=0
centre 2, – 12 and
radius 4 10
Class Exercise - 4
Find the equation of a circle which passes through origin, cut positive x and positive y-axis at 4 and 6 units from origin respectively
Solution:
O(0,0) B(4,0)
C(0,6)Diameter,asBOC=900
A (0, 6), B(4, 0)are the end points of diameter.
Equation of circle is: (x–0)(x–4)+(y–0)(y–6)=0
or, x2+y2–4x–6y=0
Class Exercise - 5
Find the equation of circle, the end-points of whose diameter are the centers of the circles x2+y2+6x– 14y–1=0andx2+y2–4x+10y–2=0.
Solution:
2 21
1
S x y 6x 14y 1 0
C : 3,7
2 22
2
S x y 4x 10y 2 0
C : 2, 5
.C1(-3,7)
S1=0. C2(2,-5)
S2=0
Diameter
(x+3)(x–2)+(y–7)(y+5)=0
or, x2+y2+x–2y–41=0
Class Exercise - 6
Find the equation of a circle passes through the points(2,3),(0,–1) and center lies on the line 3x–4y+1=0.
Solution:
Let the equation of the circle be x2+y2+2gx+2fy+c=0 ...(i)
Centre : (–g,–f) -3g+4f+1=0 ...(ii)
As(2,3) lies on (i) 13+4g+6f+c= 0...(iii)
As(0,–1) lies on (i) 1–2f+c=0 ...(iv) 1
f 1 c2
1
g 3 2c3
Putting g and f in (iii)
4 1
13 3 2c 6 1 c c 03 2
c=–3; f=–1; g=–1
Solution Cont.
x2+y2+2gx+2fy+c=0…..(i)
c=–3; f=–1; g=–1
Equation of the circle is: x2+y2–2x–2y–3=0
.
Class Exercise - 7
Find the equation of the circle which passes through (1, 1), (2, 2) and radius is 1.
Solution:
A(1,1)
B(2,2)
C(h,k)As it passes through P(1,1)
(h–1)2+(k–1)2=1r=1
h2+k2–2h–2k+1=0 ...(i)
As it passes through Q (2,2) (h–2)2+(k–2)2=1
h2+k2–4h–4k+7=0 ...(ii)
(i)—(ii) h+k=3 k=3–h
Solution Cont.
h2+k2–2h–2k+1=0 ...(i)
h2+k2–4h–4k+7=0 ...(ii)
k=3–h
From (i): h2+(3–h)2–2h–2(3–h)+1=0
2h2–6h+4=0 h=1,2
for h=1,k=2; for h=2,k=1
Equation of the circle is
(x–1)2+(y–2)2=1
(x–2)2+(y–1)2=1
h2–3h+2=0
Class Exercise - 8
If one end of diameter of circle x2+y2–4x–6y+11=0 is (8,4). Find the co-ordinate of other end?
Solution:
Center of the circle is (2,3)
Let the other end be (a,b)
a 82 a 4
2
b 43 b 2
2
other end is (–4, 2)
Class Exercise - 9
Find the equation of circle touching x+y=2 and 2x+2y=3 and passing through (1,1)
Solution
x+y=2
2x+2y=3
P(1,1)
QObserve: 1. x+y=2 and 2x+2y=3 are the parallel lines.
Diameter of the circle= Distance between the parallel lines.
2. Point (1,1) lies on x+y=2Line through P and perpendicular to the parallel line is diameter of circle It intersect 2x+2y=3 at
diametrically opposite point.
Solution Cont.
x+y=2
2x+2y=3
P(1,1)
QSlope of the parallel lines =–1
PQ is the diameter of the circle,which is r to the given lines.
Slope of PQ = 1
y 1Equation of PQ : 1
x 1
y = x
Q intersection of
x=y
2x+2y=3 x=y
3 3,
4 4
Solution Cont.
x+y=2
2x+2y=3
P(1,1)
Q3 3
,4 4
End-Points of the diameter of the circles are P(1,1) and Q(3/4,3/4)
Equation of the circle is
(x–1)(x–3/4)+(y–1)(y–3/4)=0
(x–1)(4x–3)+(y–1)(4y–3) = 0
4x2–7x+4y2–7y+6=0
Class Exercise - 10
If the equation of a circle is ax2+(2a–3)y2–4x–1=0 then the centre is
(a)(2,0) (b)(2/3,0) (c) (-2/3,0) (d) None of these
Solution:
If ax2+(2a–3)y2–4x–1=0 represents circle
a=2a–3 a=3
Equation of circle is: 3x2+3y2–4x–1=0
x2+y2–(4/3)x–1=0
Center: (2/3,0)
Thank you