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Question

Find the equation to the circle which passes through the points (1, - 2) and (4, - 3) and which has its centre on the straight line 3x + 4y = 7.

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Solution

Let the center of the circle be (h,k)
Since the center is equidistant from all the points on the circle,
(h1)2+(k+2)2=(h4)2+(k+3)2
h22h+1+k2+4k+4=h28h+16+k2+6k+9
6h2k20=0 or 3hk10=0 ........(1)
The center also satisfies 3h+4k7=0
Finding the intersection of these lines, we get

3hk10 = 3h+4k7

Put value in equation (1)

5k+3=0 or k=35
3h=1035=475 or h=4715
The radius will be (47151)2+(35+2)2=(3215)2+(75)2=1024+441225=1465225
The circle equation is therefore (x4715)2+(y+35)2=29345

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