04
CCincle
Topie : Diteremination of the equatíon of a eincle.
*Detarmtnatton of 4he equation from the center Hey I
CÉnale
ahd adius’
go
(we can understand it throuah Some examples)
Math
Detapmine the eqvation of a airale wose eenten is (-4,3) and
nadius is 5.
Solvtion: It is aiven that, (hkK)’ (-45 3)
And
a equatton of a oínele i-h)+ (y-K)=
Fonmula of a
fopmulaat4)+(9-3j=g2
So,If we put he values in th
of the airaole:
This is the requiped equation
(K+4+(y-3)= 62w
questton, +Hhís J+-K)=n)
Zememben, If he nadius ís aíven in the
fonmula of he oinele will be easy fon us to vse
, **If he denten af one cinole and the eguation of
the othen einglels
Qenter arre given:
MATH
Detenmine the equaton of the ainale whase center is at the point
(4,5) and which passes thraOug h he oenten of the inele given
by p + y + 4 -Gy =0.
Solution:
The oemten of Hhe finst incle, C,= (45)
Noà we will find the cente of he se cond afrele from
its equation We Kno wtyt22+2fy+o=0 ’¢a,-f’centert
S0, ty+4xt-6y=O
Oro, x'tyt2x2XX + 2x(-3) xyto =0
a eíncle
companima with the stan dand e guatton ot
By
4he centen f this cinle (s found to be:-233)
Henc, Ca= the dístanoe betweem Hhe aenteng ot the tw
eínele s. =N4t2)+ (5-3)2
the
ce is actually the nadíus of
wEquathon:-4)(y-= 4 finst cínele]
, fon findíng the distan¢e between two point
NOTE The fonmula
is: pX2y2 y1’These
ane the
Nexa-x +(92-90 two points.
co0n dinates of he
he eguation ot one cima le
lt toO Cinele have 4he same eenten and
is give»
MATH
that is concen tnic witn +y 4x+59 +9 =o and
and passes
The einole
Hhnough he point (2-1) has he equation ís 2
Solution: y 4 n t 5y t90>
lenten obtained trom thís equatHon
(-9,-f)(2,-5/2)
the coefficents of and y oy(-2)hwe g +
TricksBy dividing
the centen
"C;ea== (2-2)+ (-5/2 +1)
/4
e4= (2,5/2)
final equaton -2)+ (y+5/2) =A Fnom thfis
e qu atio)
CCincle
Topie : Diteremination of the equatíon of a eincle.
*Detarmtnatton of 4he equation from the center Hey I
CÉnale
ahd adius’
go
(we can understand it throuah Some examples)
Math
Detapmine the eqvation of a airale wose eenten is (-4,3) and
nadius is 5.
Solvtion: It is aiven that, (hkK)’ (-45 3)
And
a equatton of a oínele i-h)+ (y-K)=
Fonmula of a
fopmulaat4)+(9-3j=g2
So,If we put he values in th
of the airaole:
This is the requiped equation
(K+4+(y-3)= 62w
questton, +Hhís J+-K)=n)
Zememben, If he nadius ís aíven in the
fonmula of he oinele will be easy fon us to vse
, **If he denten af one cinole and the eguation of
the othen einglels
Qenter arre given:
MATH
Detenmine the equaton of the ainale whase center is at the point
(4,5) and which passes thraOug h he oenten of the inele given
by p + y + 4 -Gy =0.
Solution:
The oemten of Hhe finst incle, C,= (45)
Noà we will find the cente of he se cond afrele from
its equation We Kno wtyt22+2fy+o=0 ’¢a,-f’centert
S0, ty+4xt-6y=O
Oro, x'tyt2x2XX + 2x(-3) xyto =0
a eíncle
companima with the stan dand e guatton ot
By
4he centen f this cinle (s found to be:-233)
Henc, Ca= the dístanoe betweem Hhe aenteng ot the tw
eínele s. =N4t2)+ (5-3)2
the
ce is actually the nadíus of
wEquathon:-4)(y-= 4 finst cínele]
, fon findíng the distan¢e between two point
NOTE The fonmula
is: pX2y2 y1’These
ane the
Nexa-x +(92-90 two points.
co0n dinates of he
he eguation ot one cima le
lt toO Cinele have 4he same eenten and
is give»
MATH
that is concen tnic witn +y 4x+59 +9 =o and
and passes
The einole
Hhnough he point (2-1) has he equation ís 2
Solution: y 4 n t 5y t90>
lenten obtained trom thís equatHon
(-9,-f)(2,-5/2)
the coefficents of and y oy(-2)hwe g +
TricksBy dividing
the centen
"C;ea== (2-2)+ (-5/2 +1)
/4
e4= (2,5/2)
final equaton -2)+ (y+5/2) =A Fnom thfis
e qu atio)
