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Analytical Geometry

Cool goIITian

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28 Oct 2007 22:22:09 IST

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circle question

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Obtain the equation of the circle orthogonal to both the circles x^2+y^2+3x-5y+6=0 & 4x^2+4y^2-28x+29=0 & whose centre lies on the line 3x+4y+1=0

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Bipin Dubey

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Joined: 23 Jan 2007

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28 Oct 2007 22:37:58 IST

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For two circles : x²+y²+2g₁x+2f₁y+c = 0 and x²+y²+2g₂x+2f₂y+c = 0

Condition of orthogonality : 2g₁g₂ + 2f₁f₂ = c₁ + c₂

Let the circle be x²+y²+2gx+2fy+c = 0 whose centre is (-g,-f).

Given circles are : x² + y² + 3x - 5y + 6 = 0 & x²+ y²- 7x + 29/4 = 0

Applying orthogonality conditions with the two given circles :

3g - 5f = 6 + c............(1)

-7g + 0 = 29/4 + c...............(2)

Centre(-g,-f) lies on 3x+4y+1=0 :
This gives 3(-g)+4(-f)+1 = 0
3g + 4f = 1.............................(3)

Solving these three equations we get : g=0, f = 1/4, c = -29/4

So the equation of circle is :

x² + y² + 2(0)x + 2(1/4)y - 29/4 = 0

4x² + 4y² + 2y - 29 = 0

Swati

Blazing goIITian

Joined: 3 Aug 2007

Posts: 415

28 Oct 2007 22:50:51 IST

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Eq of 1st circle x^2 + y^2 + 3x - 5y + 6 = 0 ( 1 )

Eq of 2nd circle 4x^2 + 4y^2 - 28x + 29 = 0

or x^2 + y^2 - 7x + 29/4 = 0 ( 2 )

Let the eq of required circle be x^2+y^2+2gx+2fy+c = 0 ( 3 )

Since circle (3) cuts circles (1) & ( 2 ) orthogonally

then 2g( 3/2 ) + 2f( - 5/2 ) = c + 6

3g - 5f = c + 6 ( 4 )

and 2g( - 7/2 ) + 2f.0 = c + 29/4

- 7g = c + 29/4 ( 5 )

By subtracting we get,

10g - 5f = - 5/4

40g - 20f = - 5 ( 6 )

Given line is 3x + 4y + 1 = 0 ( 7 )

Since centre ( - g, - f ) of circle ( 3 ) lies on ( 7 )

So, - 3g - 4f = - 1 ( 8 )

Solving ( 6 ) & ( 8 ), we get

g = 0 & f = 1/4

therefore from ( 5 ), c = - 29/4

therefore from ( 3 ), required circle is

x^2 + y^2 + y/2 - 29/4 = 0

4( x^2 + y^2) + 2y - 29 = 0

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