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

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.

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

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.

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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