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Question

The equation of the circle which passes through the points (2, 3) and (4, 5) and the centre lies on the straight line y - 4x + 3 = 0, is

A
x2+y2+4x10y+25=0
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B
x2+y24x10y+25=0
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C
x2+y24x10y+16=0
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D
x2+y214y+8=0
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Solution

The correct option is B x2+y24x10y+25=0
First find the centre. Let centre be (h, k), then
(h2)2+(k3)2=(h4)2+(k5)2 ….(i)
and k - 4h + 3 = 0 ….(ii)
From (i), we get -4h - 6k + 8h + 10k = 16 + 25 - 4 - 9 or 4h + 4k - 28 = 0 or h + k - 7 = 0 ....(iii)
From (iii) and (ii), we get (h, k) as (2, 5). Hence centre is (2, 5) and radius is 2. Now find the equation of circle.

Trick : Obviously, circle x^{2} + y^{2} - 4x - 10y + 25 = 0 passes through (2, 3) and (4, 5).

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