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in Circles by (15 points)

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Let the centre of the circle be P(x,y) and the radius be r.

A(1,-6) B(7,0) C(-1,4) and D(a,2) are equidistant from P.

Therefore, AP​​​​​2=CP​​​​​2

By distance formula;

(x-1)2 + (y+6)2=(x+1)2+(y-4)2

This gives x-5y-5=0 -------(i)

Also AP​​​​​2=BP​​​​​2: 

(x-1)2 + (y+6)= (x-7)2 + y​​​​​​2

This gives: x+y-1=0------(ii)

Subtracting i from ii

We get 6y+4=0 or y=-2/3

Putting this in (ii)

We get x=5/3

Now, AP​​​​​2​= DP​​​​​2

​​(x-1)2 + (y+6)= (x-a)+(y-2)2

Putting values of x and y

We get: 3a2-10a-57=0

Applying quadratic formula, we get a= 19/3 or -3.

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