  Question: Locus of the centres of the circle which cut the circles x2 +y2+4x-6y+9=0 and x2+y2-4x+6y +4=0

Posted by: Gauri on 30.10.2019 Satyendra V. from Bhopal
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In order to find the locus of the centres of the circles intersecting the given circles we subtract the equations of the two circles.On subtracting them we get 8x-12y+5=0.This is a linear equation in x and y.Therefore locus is a straight line.  Aritra D. from Kolkata
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Te locus of the centres of the circles can be found out by subtracting the second circlefrom the first.

Thus, locus is: 8x - 12y + 5 = 0

S
Selina V. from Kochi
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Let (h,k) be the centre of the circle which cuts the two given circles.

since (h,k) passes through x2 + y2 +4x-6y+9 =0 , we have h2+k2+4h-6k+9=0 ....................(1)

since (h,k) passes through x2+y2-4x+6y+4=0 , we have h2+k2-4h+6k+4=0..........................(2)

substrating the two equations above we have , 8h-12k+5=0

simply replacing (h,k) by (x,y)  we have 8x-12y+5 =0 , which is a straight line.This is the required equation of the locus of the centres of the circle.

M
Malay K. from Kolkata
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x2+y2+4x-6y+9 = 0 and x2+y2-4x+6y+4 = 0 are the equations of two circles, Substracting we get 8x -12 y +5 =0; This is the required equation of the locus of the centres of the circle which cuts the circles give above.

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